Subnetting question

neolight90

so I have managed to go 4 months without asking a question but I have to know.

I got this practice question and following my way of doing subnets I got the wrong answer and I am racking my brain on this one as to why.

You have subnetted the 192.168.14.0 network into 8 subnets. What is the valid host range of the third subnet?

so my working out went like this so mask is 255.255.255.0

8 networks 8=4 bits as its 128 64 32 16 8 4 2 1 <<<<<and 8 is 4 in from the right so 4 bits

so mask is 255.255.255.0 change to binary and add the bits which is 4 from the left 11111111.11111111.11111111.11110000

that makes mask 255.255.255.240 which is 16 in binary so increment should be 16 right but when I saw the answer it says it should be 224 with 32 increment to get that would be 3 bits how do you get 3 bits when 8 is 4 bits

Am I just reading this question wrong. I can subnet anything normally and this is the first time in months I have been stuck.

Thanks

TheNewITGuy

How many bits from /24 must we borrow to create 8 subnets?

2^3 = 8
2^5 = 32 -2 = 30 hosts per subnet

192.168.14.0/27


range
192.168.14.1 - 192.168.14.30
192.168.14.33 - 192.168.14.62
192.168.14.65 - 192.168.14.94 < -- third subnet


Your math was off. you used 4 bits to equal 8 instead of 3

shabeerm

Here block size is 32 and mask will be 255.255.255.224 So Answer will 192.168.14.65-192.168.14.94