Is the following true or false? Subnet the IP Address 203.10.93.0 /24 - ProProfs Discuss

# Is the following true or false?

Subnet the IP Address 203.10.93.0 /24 into 30 Subnets. 203.10.93.30 a valid Host ID after subnetting.

This question is part of ccna subnetting quiz

Asked by Darwyn, Last updated: Jan 19, 2021

#### basantarana

Basantarana

Yes.For 30 subnet we need to borrow 5 bits from the host portion i.e 2^5=32 which stisfies our requirement of 30 subnets So /29 or 255.255.255.248 is the required subnet.

As Block Size of the subnet is 8 (256-248) ,its subnet ranges will be in the dfference of 8 i.e 0-7,8-15.......24-31.....32-39.......247-255.

So,Yes..........203.10.93.30 is a valid id host in the fourth subnet.

We can't subnet a network in 30, however it's possible to subnet it in 32. In this case, considering it is an IP Address class C (/24), we have to borrow 5 bitsin the 4th octet, that is, the IP Address203.10.93.0 /24 will turninto203.10.93.0 /29 and mask 255.255.255.248. Dividing the network 203.10.93.0 in 32 it means we will have 8 IP'sper subnet (only 6 of them are valids because the first IP is the Network ID and the last IP is the Broadcast ID, we can't use them).Now let's check out the subnets we'vegot.

1- NID 203.10.93.0
BID 203.10.93.8

2- NID 203.10.93.9
BID 203.10.93.16

3- NID 203.10.93.17
BID 203.10.93.24

4- NID 203.10.93.25
BID 203.10.93.32 And it goes on until 255.

As we can see, the IP 203.10.93.30 is in the 4th subnet and it is not a NID and even a BID, so it is a valid IP address.

Yes, but I disagree with a couple of the explanations on here - each subnet would only hold 8 (or 6 useable) hosts. Borrowing 5 bits allows a further 32 subnets, but only 8 hosts (again, 6 useable).

The host in question would reside on the 4th subnet (it would actually be the last assignable host).

#### mehmood.usman

Yes
Because range of first useable subnets is (203.10.93.1/29 to 203.10.93.30/29)

#### Priyankagehu

Yes would be the ans.2^5=32 that means 5 bit borrowed frm the host part and each ntwork will have 30 hosts.
the above needed ip would be in first network itself.
203.10.93.0/29 to 203.10.93.31/29

#### John Smith

John Smith

Yes
203.10.93.0 Need 30 Networks 128 64 32 16 8 4 2 1 0 0 0 1 1 1 1 0 = 30 in binary Borrow 5 Bits for # of Networks New Subnet Mask = 255.255.255.11111000 (248) or /29 ^ Increment = 8 2^5 = # of Networks = 32 2^3-2 = # of Hosts per Network = 6 First 5 IP Ranges 223.10.93.0 - 223.10.93.7 93.8 93.15 93.16 93.23 93.24 93.31 93.32