For a class C network if IP address of a computer is 200.99.39.112 and subnet mask is 255.255.255.224 the decimal value of last octet of last host of sixth subnet is ____________.

Subnet mask is 32 bit dotted decimal represented number created by setting all network bit to $1$ and all host id bit to $0$.

so here subnet mask =$11111111.11111111.11111111.11100000$ which implies we have $27$ bit as network bit and $5$ bit as host id.

Given IP address belongs to class C network which has $24$ bit as network id and $8$ bit as host id .But subnet mask we have $27$ bit bit as network id and $5$ bit for host id . So it implies we use $(27-24)=3$ bit for subnetting which gives us $2^{3}=8$ subnet.

Now we need network IP address for the given IP address for that we need to do bit wise and with IP address and Subnet mask.

$200.99.39.112$

$255.255.255.224$

$200.99.39.96$

So if we expand the last octet in binary we see

$01100000$ ---- It represent 4^{th} subnet network id as subnet bit is $(011)_{2}$=$3_{10}$ , It is 4^{th} subnet as first subnet is $(000)_{2}$=$0$ ,second subnet is $(001)_{2}$=1 so on.

Now we need 6^{th} subnet for that subnet bits would be 101.

So the network id of the 6^{th} subnet is

$200.99.39.160$ -----[$160_{10}$=$10100000_{2}$]

Last assignable ip address for this subnet is when all the host id is 1 except the last bit so that is should not be broadcast address for this subnet .

so The last assignable IP address for the 6^{th} subnet is

$200.99.39.190$.

So the decimal value of last octet of last host of sixth subnet is $190$.