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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
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 octetQuote:
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
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..
Oh....I'm just going to let this one go and watch the fur fly for awhile. Should be fun.
I'm still gonna call ya an idiot.. because I can..
let's take a look at a North Carolina State University practice question...
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...Quote:
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 1’s followed by a group of 0’s. In this case, the binary representation of this incorrect mask is:
1111 1111 1111 1111 0001 1111 0000 0000
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..
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
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.
Yeah, or he forgot letter E -- None of the above.
:)
This is the exact reason why they invented IP subnet calculators....yes I cheat, whatever it takes to get the job done, right?
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.
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...
Eh, I dredged this up from the depths... Almost makes we wish I never posted a tutorial--because now I have a bit "Tutorials:1" near my name and it just looks forlorn.
Now I’m confused.
If you take the tutorial from TERR and apply it, you have
IP1 10000000.00001000.10000001.00101011 (128.8.129.43)
IP2 10000000.00001000.10100001.00110111 (128.8.161.55)
mask 11111111 .11111111.00011111.00000000 (255.255.31.0 )
10000000.00001000.---00001.-------- (128.8.129.43)
10000000.00001000.---00001.-------- (128.8.161.55)
Couldn’t they be on the same network?? Is it possible??
well well well !!! computers are idiots .not WE!!!
coming to details of question, this question is being asked in GATE(graduate aptitude test in engineering) in india and there IS an answer to this one which i dunno.
i had worked on subnets problems like given IP address and 'n' systems in network etc etc.....
but this Q is not asked directly.
I think 'D' is correct answer. but i dont have solution!
Ahem, someone needs to go review RFCs... Netmasks MUST be continuous "1" bits.
Want to try prove me wrong?: Tell me the network and host addresses of 63.146.109.212 if it has a subnet mask of 255.255.31.0. Good luck.
Ammo
SolarWinds makes a nice subnet calculator. It's the one I use all the time although I am going to check out HTRegz's.
http://support.solarwinds.net/update...stomerFree.cfm
I would love to see someone give me an answer as to how N.N.31.x could be a legit subnetmask. Maybe this question was meant to leave blank since there is NO real answer??? ;)
Agreeing with ammo, the network side of the subnetmask has to be all '1' bits, and the host side of the subnetmask are all '0's.
And to go a little further into ammo's discussion, to show you proof that there is no such thing as a N.N.31.x subnetmask, look at RFC 1878