Three subnet bits only will give you 6 subnets (2^3-2) so you have to use 4
bits, hence the 240 (11110000).
""Anton Chu"" wrote in message
news:200110052001.QAA31118@xxxxxxxxxxxxxxxxx
> You're going to have convert the IP and subnet to binary notation thus
> getting:
> 201.110.65.0 -> 11001001.01101110.01000001.00000000
> 255.255.255.240 -> 11111111.11111111.11111111.11110000
> Mulitply through ---------------------------------------------------
> 11001001.01101110.01000001.11110000
>
> To detemine the number hosts, you count the trailing zeros which is 4.
> # of hosts = 2*nth power - 2
> nth power = number of trailing zeros
> total hosts = 14
>
> Maybe someone can explain how getting 7 subnets is derived.
>
> Anton
>
> ----- Original Message -----
> From:
> To:
> Sent: Friday, October 05, 2001 12:17 PM
> Subject: Q again... [1:6342]
>
>
> > Hello guys,
> >
> > Back with my subnetting questions and I need your help again.
> >
> > Please explain how the answer to this question was arrived at.
> >
> > 1. Given the class c address 201.110.65.0 design a subnet mask to
support
> 7
> > subnets with 14 hosts per subnet.
> >
> > a) 255.255.255.240
> > b) 255.255.255.0
> > c) 255.255.255.248
> > d) 255.255.255.224
> >
> > Boson test says the answer is A. (255.255.255.240). Pls explain.
> >
> > Much Appreciated.
> > Tee.
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