question on subnet mask
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Thread: question on subnet mask

  1. #1
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    question on subnet mask

    this is question asked in one of competitive exams

    Q)the subnet mask for a particular network is 255.255.31.0. which of the following pairs of IP addresses could belong to this network?

    A) 172.57.88.62 and 172.56.87.233
    B) 10.35.28.2 and 10.35.29.4
    c) 191.203.31.87 and 191.234.31.88
    D) 128.8.129.43 and 128.8.161.55

    if anyone knows solution please let me know
    thanks in advance

  2. #2
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    Originally posted here by NeuTron
    do your homework

    *hint* its a class b address using a wildcard. Only one of the answers could possibly fit the description, look closely
    Man are you an idiot??? Of course only one answer could fit the description that's how multiple choice questions work...Also you have no clue that it's a class B.. It could be a class A with more than 8 bits borrowed.... The entire problem with this question is that 255.255.31.0 is not a valid subnet mask. The entire idea of subnetting is that you are borrowing bits. to get 31 you would have to have 00011111..... this would be valid for an ACL but not for a Subnet Mask.... There are only 9 Valid Masks for each octet
    11111111 - 255
    11111110 - 254
    11111100 - 252
    11111000 - 248
    11110000 - 240
    11100000 - 224
    11000000 - 192
    10000000 - 128
    00000000 - 0
    All octets of the SM must be made up of one of those numbers.. otherwise it isn't a subnet mask..


    Now I could be wrong..... (I suppose the possibility of wacky theories is out there) but after 4 years of networking theory and 2 years of tutoring/teaching it.. I feel that I have it down pretty good..
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  3. #3
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    Oh....I'm just going to let this one go and watch the fur fly for awhile. Should be fun.

  4. #4
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    I'm still gonna call ya an idiot.. because I can..

    let's take a look at a North Carolina State University practice question...

    Source: http://ecommerce.ncsu.edu/technology...ructure-a.html
    What is wrong with the following mask: 255.255.31.0?
    ANSWER: If the mask is converted to binary, It should be group of 1ís followed by a group of 0ís. In this case, the binary representation of this incorrect mask is:
    1111 1111 1111 1111 0001 1111 0000 0000
    Now if this is a Wildcard mask, then it is not a subnet mask... again negating the original question... because well none of those would ever match the same ACL...

    I have spent over an hour reading the cisco curriculum and searching google and I have yet to see anything on Subnetting Wildcards.... Subnetting is a pretty standard practice and I don't see them changing it to allow for wildcards.. but feel free to post links and prove me wrong.. you may even get an apology out of me.. but I doubt it since it isn't my way..
    IT Blog: .:Computer Defense:.
    PnCHd (Pronounced Pinched): Acronym - Point 'n Click Hacked. As in: "That website was pinched" or "The skiddie pinched my computer because I forgot to patch".

  5. #5
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    If I may ask, what competitive exam did this question come out of? I would be interested, if HTRegz is correct that this question is faulted, why this could be on a competitive exam...

    Regards...
    ~AciD
    [shadow]There is no right and wrong, only fun and boring...
    Formatting my server because someone hacked into it sounds pretty boring to me...
    That\'s why it\'s all about AntiOnline.com!
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  6. #6
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    Its not IF he is correct....its He IS correct. Theres no such thing as Subnetmask Wildcards.....and a .31 would end up like a wildcard bit.

    Maybe they meant 25.255.224.0 and just did a mis-typo.

    if that is the case then

    A wouldnt be the answer because obviously the 2nd octet isnt the same.
    C wouldnt be the answer because the same reason for answer A.
    D would be the incorrect answer because they are not in the same /19 subnet

    B would be the correct answer. because the first two octets have to stay the same since the subnetmask would be
    255.255.x.x

    Since it would be a .224 in the 3rd octet. your options would be
    10.35.0.0 - 10.35.31.255
    10.35.32.0 - 10.35.63.255
    10.35.64.0 - 10.35.95.255
    10.35.96.0 - 10.35.127.255
    10.35.128.0 - 10.35.159.255
    10.35.160.0 - 10.35.191.255
    10.35.192.0 - 10.35.223.255
    10.35.224.0 - 10.35.255.255

    With 10.35.28.2 and 10.35.29.4 the .28 and .29 fall under the same subnet.
    Os1LaYr5

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    Master-Jedi-Pimps0r & Moderator thehorse13's Avatar
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    Yeah, or he forgot letter E -- None of the above.

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  8. #8
    Priapistic Monk KorpDeath's Avatar
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    This is the exact reason why they invented IP subnet calculators....yes I cheat, whatever it takes to get the job done, right?
    Mankind have a great aversion to intellectual labor; but even supposing knowledge to be easily attainable, more people would be content to be ignorant than would take even a little trouble to acquire it.
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  9. #9
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    Then again, it could be a trick question. You can't get the answer right, but they see what people put anyway. I got a bad question when I took the N+ some years ago.
    Those who speak of what they know... find out too late that prudent si-lence was wise.
    --Madame Giry, Phantom of the Opera

  10. #10
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    KorpDeath Mentioned Subnet calculators...

    I posted this one that I coded in VB sometime ago.. but I'll post it again because it might have some relevance to those reading this thread.


    You'll have to have the VB Runtime files...
    IT Blog: .:Computer Defense:.
    PnCHd (Pronounced Pinched): Acronym - Point 'n Click Hacked. As in: "That website was pinched" or "The skiddie pinched my computer because I forgot to patch".

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