How many subnets and hosts are possible given the Mask Address and these IP addresses

Question

This is a new topic for my Teleprocessing and Network Class. Don't take me wrong, I made my research and I got partial answer to this question, but I'm not sure of what do I have is correct.

We found a number of computers with the same subnet-subnet mask (255.255.255.224) and whose IP addresses are set out below. Indicate how many subnets and nodes are possible.

192.168.1.1, 192.168.1.34, 192.168.1.67, 192.168.1.100 192.168.1.2, 192.168.1.36, 192.168.1.70, 192.168.1.104 192.168.1.3, 192.168.1.37, 192.168.1.69, 192.168.1.103 192.168.1.4, 192.168.1.40, 192.168.2.71, 192.168.2.111 192.168.2.5, 192.168.2.44

I know that from that Mask address this is a Class C network in which we can have a Maximum of 254 nodes.
Then I search in the Cisco website that:

• Number of bits: 3 (because we took 3 bits to make that exact mask address)
• Number of Subnet:8 (becausd of those 3 bits)
• Number of Host per Subnet: 30

Now, what is the meaning of those IP adresses below,
Did I answered the question correctly?
Can anyone explain me in detail?

Answer 1

CIDR (Classless Inter-Domain Routing) gives us an opportunity to subnet classful networks and therefore use masks different than /8 /16 or /24 (You can find some information about VLSM (Variable Length Subnet Mask) ).

Subnet mask 255.255.255.224 (/27) as you said means we have total 32 addresses in a network (network address, broadcast address + 30 usable host addresses). According to your IP's list you should find (as far as I properly understood your task) all /27 subnets which covers aforementioned IPs. For instance:

network 192.168.1.0/27 (addresses 192.168.1.0 - 192.168.1.31) 
    "eats" 192.168.1.1, 192.168.1.2, 192.168.1.3 and 192.168.1.4 from your list
network 192.168.1.32/27 (addresses 192.168.1.32 - 192.168.1.63)
    "eats" 192.168.1.34, 192.168.1.36 ...
...
...
...
and so on