Here's an oddball test question

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

pearson question.jpg

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RouteMyPacket

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. : )

networker050184

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.

Jon_Cisco

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

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

I doubt you'd see something like this on the CCNA honestly.

theodoxa

BGraves wrote: »

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.

theodoxa

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.