Here's an oddball test question
mikeybinec
Member Posts: 484 ■■■□□□□□□□
in CCNA & CCENT
This is from Pearsonitcertification.com (They publish the exam cram books, amongst others)
I know the answer but was hoping our resident gurus can share their insight in solving this
I know the answer but was hoping our resident gurus can share their insight in solving this
Cisco NetAcad Cuyamaca College
A.S. LAN Management 2010 Grossmont College
B.S. I.T. Management 2013 National University
A.S. LAN Management 2010 Grossmont College
B.S. I.T. Management 2013 National University
Comments
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RouteMyPacket
Member Posts: 1,104
mikeybinec wrote: »This is from Pearsonitcertification.com (They publish the exam cram books, amongst others)
I know the answer but was hoping our resident gurus can share their insight in solving this
A I am guessing
Actually it's C after further staring at screen like i'm curing cancer.
10.32.0.0 - 10.32.0.1
10.32.0.2 - 10.32.0.3
10.32.0.4 - 10.32.0.5
10.32.0.6 - 10.32.0.7
10.32.0.8
and so on...ugh..that one got me at first but such is life..nice question this morning. : )Modularity and Design Simplicity:
Think of the 2:00 a.m. test—if you were awakened in the
middle of the night because of a network problem and had to figure out the
traffic flows in your network while you were half asleep, could you do it? -
networker050184
Mod Posts: 11,962 Mod
So write out all of the even numbers in binary.
0 - 00000000
2 - 00000010
4 - 00000100
........
48 - 11000000
.......
102 - 11001100
.......
224 - 11100000
If you go through the whole list what is the the common bit you could match on? The first bit (far right) is always going to be a 0 in any even number. That should get you going in the right direction.An expert is a man who has made all the mistakes which can be made. -
Jon_Cisco
Member Posts: 1,772 ■■■■■■■■□□
Ignoring the first 3 octets you need to determine what the wild card is asking for.
A ...252 x.x.x.1111 1100 first 6 bits allow anything last two bits must match
B ...254 x.x.x.1111 1110 first 7 bits allow anything last bit must match
C ...254 x.x.x.1111 1110 first 7 bits allow anything last bit must match
D ...255 x.x.x.1111 1111 all 8 bits allow anything
I think I would narrow it down to B or C.
Then look at my starting address I think I would choose C.
I don't know if I'm right but that's how I read it. -
BGraves
Member Posts: 339
Good grief, this is a CCENT/CCNA test prep question? Glad I got mine....2 1/2 years ago. (crap, just remembered I should renew this one) -
networker050184
Mod Posts: 11,962 Mod
I doubt you'd see something like this on the CCNA honestly.An expert is a man who has made all the mistakes which can be made.
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theodoxa
Member Posts: 1,340 ■■■■□□□□□□
Good grief, this is a CCENT/CCNA test prep question? Glad I got mine....2 1/2 years ago. (crap, just remembered I should renew this one)
I seriously doubt it. ACLs with interleaved zeroes is a CCIE level topic. I don't think its even touched on at the CCNP level, much less CCNA.R&S: CCENT → CCNA → CCNP → CCIE [ ]
Security: CCNA [ ]
Virtualization: VCA-DCV [ ] -
theodoxa
Member Posts: 1,340 ■■■■□□□□□□
C - BTW, B and D can be immediately eliminated as the ending digit of the subnet ID is odd and it would automatically match itself. The correct answer is C, because it ignores everything, but the final digit in the last octet. Since a 1 in the 1s place is the only way to achieve an odd number in binary,
A permit ip 10.32.0.0 0.31.255.252 any
B permit ip 10.32.0.1 0.31.255.254 any
C permit ip 10.32.0.0 0.31.255.254 any
D permit ip 10.32.0.1 0.31.255.255 any
As is typical of Pearson's practice tests, the wording is not very good. Technically, none of the answers would be correct since even numbered should mean all octets are even numbered - at least in my mind. They are obviously only referring to the final octet.R&S: CCENT → CCNA → CCNP → CCIE [ ]
Security: CCNA [ ]
Virtualization: VCA-DCV [ ]
