subnet question

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can someone please confirm the answer to this:

For the following class B address, what is the network address of the fourth usable subnet ?
162.21.122.152 /25

Thanks kindly.

 

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Its a bit early but I believe if Im reading it right you would have to break it down to a /27 to get 4 subnets:

4 network address woul be:

162.21.122.128/27
162.21.122.160/27
162.21.122.192/27
162.21.122.224/27

so the 4th would be 162.21.122.224/27 gateway .225 broadcast .255

 

This is a bit confusing to me. At first I was thinking about it the same way as the 1st answer by theITCrowd but if you are given a class B address then the normal subnet mask is /16 so 9 bits have already been borrowed. 2^9 is 512 so there are 512 subnets without taking any further bits. ???

 

The question doesn't make any sense to me either, but for different reasons.

 

HELLO

may answer is if u want 2 se the rang of the subnet u should look at the last bis on subnet mask

/25= 255.255.255.128 the rang is

162.21.122.0 ---162.21.122.128
162.21.122.129---162.21.123.0
162.21.123.1----162.21.123.128
162.21.123.129---162.21.124.0

i think 95% is the correct answer

 

Hello Fran1942

Just started ICND1 and been doing some subnetting practice at home and thought I give this a try.

162.21.122.152/25 since 162 lies between 128 - 191 => a class B address

Interesting octet in this case is 152 => 1001 1001

Applying subnet mask to 1001 1000 gives 1000 0000 (i.e. 128 ) => network increment = 256 - 128 = 128

So the subnets are as follows:

162.21.122.128
162.21.123.0
162.21.123.128
162.21.124.0
162.21.124.128
etc..
etc..

Not entirely sure what you mean by usable subnets since if I am not mistaken they are all usable are they not

Can someone double check my calculations because I really want to know the answer to this question.

 

the zero and 128 ipes r not usable

 

162.21.122.152 /25

a /25 does give a mask of 255.255.255.128

or in binary 11111111.11111111.11111111.10000000

Technical you are in the class B range, the first octect of 162 shows this as per ibar's explanation but we're dealing with a CIDR address style as defined by the /25 which actually indicates that we're using hosts in the 4th octect only.

The remainder of the host bits 2^7 gives you 128(-2) hosts (the -2 is for the network address and the broadcast address) and the ranges of:

162.21.122.0/25
162.21.122.128/25

Something isn't right with the question as you do not have a 4th useable subnet, you only have two subnets available.

So, where'd the question come from?

A good deeper explanation is here: http://www.akadia.com/services/ip_routing_on_subnets.html

 

Agreed - there is no 4th subnet, there is something wrong with the question.