Hello,
I'm from Holland and soon i will do my CCNA examination. I study very hard for it. Could somebody help me with the next question? I know the answer is C, but could somebody explain why?
what is the subnetnumber of a host with an ip address of
A. 172.16.42.0 B. 172.16.107.0 C. 172.16.208.0 D. 172.16.252.0 E. 172.16.254.0
the third byte of the subnet mark is :
128+64+32+16+8+4=252,so subnet-mark is 255.255.252.0256-252=4
the hosts of this subnet should be from 172.16.208.1-172.16.213.254
the answer is C
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You could also try going to the following site for an explaination
Write down the address in binary:
172 16 210 0 10101100 00010000 11010010 00000000then write a /22 mask in binary:
255 255 252 0 11111111 11111111 11111100 00000000finally, do a logical AND between the two, and you get:
10101100 00010000 11010000 00000000or, in the usual notation, 172.16.208.0.
If you're going for your CCNA you better learn to do these calculation quite fast.
The range is
172.16.208.0/22 - 172.16.211.255/22172.16.213.255 belongs to 172.16.212.0
Hey,
i tried to make a little diagram to help me work out the bits etc while subnetting the least significant octet etc...
i know this aint perfect and there are possibly better diagrams out there, but this seemed to help me a bit..... .....pardon the pun!
regards
Harry
Last Octet
128 64 32 16 8 4 2 1255 1 1 1 1 1 1 1 1 /32 - 20 = 1
254 1 1 1 1 1 1 1 0 /31 - 21 = 2252 1 1 1 1 1 1 0 0 /30 - 22 = 4
248 1 1 1 1 1 0 0 0 /29 - 23 = 8240 1 1 1 1 0 0 0 0 /28 - 24 = 16
224 1 1 1 0 0 0 0 0 /27 - 25 = 32192 1 1 0 0 0 0 0 0 /26 - 26 = 64
128 1 0 0 0 0 0 0 0 /25 - 27 = 128Have something to add? Share your thoughts — no account required.
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