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Subnetting Questions

May 5th, 2011 in ICND2 Go to comments

Here you will find answers to ICND 2 – Subnetting Questions

Note: If you are not sure about subnetting, please read my Subnetting tutorial.

Question 1:

Refer to the exhibit. The lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANS in Phoenix but no additional subnets?

subnetting_1

A – 10.0.0.0 /22 is subnetted, 1 subnet
D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

B – 10.0.0.0 /28 is subnetted, 1 subnet
D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

C – 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

D – 10.0.0.0 /22 is subnetted, 1 subnet
D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

E – 10.0.0.0 /28 is subnetted, 1 subnet
D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

F – 10.0.0.0 /30 is subnetted, 1 subnet
D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

Answer: D

Question 2:

Refer to the exhibit. HostA cannot ping HostB. Assuming routing is properly configured. what could be the cause of this problem?

subnetting_2

A – HostA is not on the same subnet as its default gateway.
B – The address of SwitchA is a subnet address.
C – The Fa0/0 interface on RouterA is on a subnet that can’t be used.
D – The serial interfaces of the routers are not on the same subnet.
E – The Fa0/0 interface on Routers is using a broadcast address.

Answer: D

Question 3:

Refer to the exhibit. The routers are running RIPv2. Which addressing scheme would satisfy the needs of this network yet waste fewest addresses?

subnetting_3

A – Network 1: 1 92.168.10.0/26
Network 2: 1 92.168.10.64/26
Network 3: 1 92.168.10.128/26
Serial link 1: 192.168.20.0/24
Serial link 2: 192.168.30.0/24

B – Network 1: 1 92.168.10.0/26
Network 2: 1 92.168.10.64/28
Network 3: 1 92.168.10.80/29
Serial link 1: 192.168.10.88/30
Serial link 2: 192.168.10.96/30

C – Network 1: 1 92.168.10.0/26
Network 2: 1 92.168.10.64/27
Network 3: 1 92.168.10.96/28
Serial link 1: 192.168.10.112/30
Serial link 2: 192.168.10.116/30

D – Network 1: 1 92.168.10.0/27
Network 2: 1 92.168.10.64/28
Network 3: 1 92.168.10.96/29
Serial link 1: 192.168.10.112/30
Serial link 2: 192.168.10.116/30

Answer: C

Comments
  1. Ahsan
    August 19th, 2010

    I do not unerstand Q1 that how did you narrow down the right answer, could anyone help in this regard plzzz?

  2. fenderplus
    August 19th, 2010

    To Ahsan:
    Since 3 octet is the only diference, put this octec first in binary to understand which are the bits are the same on that octec:

    10.4.0.0=10.4.0000 0000.0
    10.4.1.0=10.4.0000 0001.0
    10.4.2.0=10.4.0000 0010.0
    10.4.3.0=10.4.0000 0011.0

    As you can see from left to right we have 22 bits that are the same 8(1st octec)+8(2rd octect)+6(3th octect) = that makes /22 the summary address.

    The summary ip therefore would be:

    10.4.0.0/22

    with mask=255.255.11111100.0 = 255.255.252.0

  3. Crono
    September 12th, 2010

    Can someone help out with Q1 I’m fine with the autosummarization of the network i am just unclear on how the eigrp summary route was determined

  4. Anonymous
    September 17th, 2010

    Auto-summarization and/or the Routing protocol are not really relevant in this question.
    The bottom line is 10.4.0.0/22 encompasses the 4 LAN networks attached to the PHOENIX Router, that the CENTRAL Router needs to know about.

  5. Paul
    October 1st, 2010

    I understan that 10.4.0.0/22 encompasses the 4 LAN Networks attached to the PHOENIX router. However the answer is very confusing:

    D – 10.0.0.0 /22 is subnetted, 1 subnet
    D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

    There is no 10.0.0.0/22 network in the question. And the secod line 10.4.0.0 has no mention of the subnet mask. So the ip address is on the second line, the subnet mask is on the first line and I don’t know what the point of 10.0.0.0 is. Why is the route summary written like this?

  6. Kent
    October 5th, 2010

    I agree with Paul, choices are not straightforward.

  7. amit
    October 11th, 2010

    10.0.0.0 is the default ip of Class A that is why it is written.

  8. gillman
    October 26th, 2010

    all these questions and the labs are on pass4sure does anyone use pass4sure and if so what do you think??

  9. gencis
    November 8th, 2010

    ” gillman
    October 26th, 2010

    all these questions and the labs are on pass4sure does anyone use pass4sure and if so what do you think?? ”

    Hey i don’t really trust it fully cause there are a lot of mistakes , belive me 9tut is the right place to be … i am going for my icnd2 very soon , good luck.

  10. K.KUMARASAMY
    December 24th, 2010

    I do not unerstand Q2 that how did you narrow down the right answer, could anyone help in this regard plzzz?

  11. Jeff H.
    January 16th, 2011

    The question gives /27 as the netmask for the serial connections. If that’s the case the host ranges is incremented by 32. Thus the ranges of valid (usable) hosts will look like this:

    1st Range: 192.168.1.1 – 192.168.1.31
    Second Range: 192.168.1.33 – 192.168.1.63 (Router A resides here)
    Third Range: 192.168.1.65 – 192.168.1.91 (Router B resides here)

  12. Jeff H.
    January 16th, 2011

    Question 1. It wants you to summarize the routes so you don’t have to put each route into central’s routing table or waste routes by being to broad by not summarizing.

    Say if another branch location popped up in Oklahoma and wanted to use 10.4.4.0/24. Well, it couldn’t. Even though the 10.4.4.0/24 isn’t being used, the central router will still route packets destined for that space to phoenix.

    To solve:

    First we know all the subnets in Phoenix are class B… /16

    So now we know the 3rd octet can be manipulated to reach our goal.

    10.4.(3).0 3rd octet 00000011 = 3. So, we only need two bits for the host in the 3rd octet.
    the other 6 bits will be used for the networking portion.

    11111111. 11111111. 11111100 00000000 = 255.255.252.0 = /22

  13. Jeff H.
    January 16th, 2011

    ooops, i meant class A for the last one :)

  14. *Vash*
    March 15th, 2011

    thank you Jeff!
    Have you recently passed the ICND2, or is prepping?

  15. Mo
    March 19th, 2011

    9tut can you please help me understand question 1.

  16. *Vash*
    March 29th, 2011

    Mo, Pay close attention to the following portion of this question…..
    Which partial routing table entry in the Central router represents a “route summary” that represents the LANS in Phoenix but no additional subnets?

    when you see “ROUTE SUMMARY, SUMMARIZED, you know it’s going to be VLSM!
    I would find your nearest download file/Bookstore and practice VLSM subnetting….

    Gravy!
    Hope this helps…
    Vash

  17. Snoop
    May 11th, 2011

    Are these questions actually on the exam or are they just guides of what you might be asked.

  18. ZR1
    May 25th, 2011

    These questions are very close to what is on the exam. The subnets and node names, and so forth will be different, but the basic idea is the same. I would not go through 9tut and memorize the answers if that’s what you are looking for.

  19. DJ
    July 6th, 2011

    ICND2 Failed! 703
    i m pretty sure that question 2 was asked about a month ago..
    too bad i could not get time to comple the exam..
    the questiones are getting bigger and bigger …
    And 9tut is becoming less reliable when it comes to rare exams like ICND2 when compaired to CCNA as a whole.. :(

  20. DJ
    July 6th, 2011

    i say less reliable cos only 2 or 3 questions out of the whole ICND2 website was on the real exam..

  21. siddy
    August 18th, 2011

    can someone explain the answer to the question 3

  22. xallax
    August 18th, 2011

    @siddy
    best practice states that the bigger subnets should be placed near the start of the address space.

    network1 has 50 hosts, a subnet of /26 would acomodate
    network2 has 20 hosts, a subnet of /27 would acomodate
    network3 has 10 hosts, a subnet of /28 would acomodate

    the serial links use 2 IPs, 2 subnets of /30 are required.

    solution:
    net1: 10.0 – 10.63
    net2: 10.64 – 10.95
    net3: 10.96 – 10.111
    serial1: 10.112 – 10.115
    serial2: 10.116 – 10.119

    do you have problems subnetting siddy? there is a subnet question generating script that comes with all my VCEs at this link:
    http://www.tinyurl.com/xallaxvce

  23. twigonometry
    August 26th, 2011

    Is anyone else concerned that in Question 2, Router A has both interfaces FA0/0 and S0/0 in the same subnet?

    ie: 192.168.1.33 / 27 and 192.168.1.62 / 27 both belong to the 192.168.1.32.0 / 27 network.

    To me this represents a problem because each interface will be listed in the route table with access to the same 192.168.1.32.0 network. The router will probably load balance between the two interfaces?

    Other than this I agree with the answer given (D).

  24. xallax
    August 26th, 2011

    @twigonometry
    yes, you’re right, both interfaces are in the same subnet.
    still, the only option that we can pick there is “The serial interfaces of the routers are not on the same subnet”.

  25. Anonymous
    November 16th, 2011

    can any one tell answer for this question?how identify network id and broadcast adress 172.16.4.2 tell with and example pleae!!

  26. Shaochieh
    November 29th, 2011

    Anonymous, do you have the mask??? You need to have the mask to figure the broadcast and network you are on.

  27. Anonymous
    December 11th, 2011

    ^^ Can we just say that 172.16.4.2 is a class B address, so default mask is 255.255.0.0

    Network ID is 172.16.00
    Broadcast is 172.16.255.255

    Is this correct?

  28. 9tut
    December 12th, 2011

    @Anonymous: To solve your question, we need to know the subnet mask of 172.16.4.2 because different subnet mask will give different network and broadcast addresses.

  29. WKC
    January 28th, 2012

    Passed ICND2 today with 944!! I didn’t have any of these on my exam today

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