""s vermill""  wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Richard Burdette wrote:
> >
> > A prep test I am using has a question for which I disagree with
> > the answer.  Here is the question.
> >
> > If I had a Class B address, what subnet mask would I use if I
> > wanted to split it into 8 class C addresses?
> >
> >
> > a.255.255.240.0
> > b.255.255.255.0
> > c.255.255.248.0
> > d.255.255.254.0
> >
> > The answer from the test is "c".
> >
> > I think the answer is not even listed; 255.255.224.0 because to
> > add eight additional subnets we need 2^3=8 bits of subnet which
> > equates to 224 of mask.  Am I right or wrong?
> >
> > Rich
> >
> >
>
> The question sucks and so do the answer choices.  Eight *addresses* per
> subnet or eight subnets?  If the former, a mask of 255.255.255.248 would
be
> required - not 255.255.248.0.  If the latter, I'd go with you (sorta).  Or
> perhaps a. if subnet zero weren't allowed for some reason (but then you'd
> have 15 subnets plus subnet zero - not eight!).
>
> Nothing you do with a subnet mask makes a class B a class C.  If it was
> class B before you subnetted it, it isn't going to fall into the class C
> range of addresses all of the sudden is it?  If by "class C" they mean a
> subnet with 254 host addresses, your answer isn't right either.  Your
> subnets (/21) would have 8190 hosts.  In that case, there isn't a valid
> answer at all because a class B subnetted for 254 hosts will have
somewhere
> in the neighborhood of 255 subnets.


CL: I disagree. 172.16.0.0/21 gives you eight subnets, each with a mask of
/24 - subnets 172.16.0.0/24 through 172.16.7.0/24, for example. Think ISP /
BGP :-)



>
> I would be wary of the "practice" test as a whole.




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