IP Address & Subnetting Questions
Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.
Question 1
Explanation
A /30 subnet means subnet mask of 255.255.255.252. But 10.2.1.3 255.255.255.252 is a broadcast IP address; only 209.165.201.2/30 is the usable IP address.
Question 2
Question 3
Explanation
This question asks about the private ranges of IPv4 addresses. The private ranges of each class of IPv4 are listed below:
Class A private IP address ranges from 10.0.0.0 to 10.255.255.255
Class B private IP address ranges from 172.16.0.0 to 172.31.255.255
Class C private IP address ranges from 192.168.0.0 to 192.168.255.255
Only the network 172.28.0.0/16 belongs to the private IP address (of class B).
Question 4
Question 5
Explanation
We need a subnet with 20 users so we need 5 bits 0 in the subnet mask as 25 – 2 = 30 > 20. Therefore the subnet mask should be /27 (with last octet is 1110 0000 in binary). The increment is 32 so the valid network address is 10.10.225.32.
Question 6
Explanation
We see the maximum number of user per floor is 29 users (Floor 2) < 32 so the best subnet mask should be 1110 0000 which allows 25 – 2 = 30 hosts per subnet.
Question 7
Explanation
Each interface on a router must be in a different network. If two interfaces are in the same network, the router will not accept it and show error when the administrator assigns it.
Question 8
Question 9
Explanation
We cannot use the subnet mask of 255.255.254.0 (/23) because it is smaller than the default subnet mask of class C (/24). If we use /23 subnet mask, in fact we are summarizing (supernet) it -> Answer B is not correct.
With the subnet mask of 255.255.255.128, we have 21 = 2 subnets only, not enough for 8 floors -> Answer D is not correct.
We need 40 users per subnet so our subnet must support at least 64 (=26) hosts (in fact 62 hosts). So the last octet of subnet mask should be 1100 0000 (with 6 bits “0”) = 192 -> The suitable subnet mask is 255.255.255.192. But this subnet mask only provides 22 = 4 subnets, which is not enough for 8 floors.
For 8 floors we need 1110 0000 (23 = 8) for the last octet of subnet mask so the subnet mask should be 255.255.255.224.
But with the subnet mask of 255.255.255.224, the host addresses per subnet is 25 – 2 = 30 so we cannot fulfill the host addresses for 40 users. We can only provide IPs for 30 users. This question said “approximately 30-40 users per floor” so “30 users per floor” is acceptable.
With “192.168.0.0 255.255.255.224” we have 8 subnets:
+ First subnet: 192.168.0.0 to 192.168.0.31
+ Second subnet: 192.168.0.32 to 192.168.0.63
+ Third subnet: 192.168.0.64 to 192.168.0.95
+ Fourth subnet: 192.168.0.96 to 192.168.0.127
+ Fifth subnet: 192.168.0.128 to 192.168.0.159
+ Sixth subnet: 192.168.0.160 to 192.168.0.191
+ Seventh subnet: 192.168.0.192 to 192.168.0.223
+ Eighth subnet: 192.168.0.224 to 192.168.0.255
Question 9 should be removed. It is not fair to assume that an engineer wouldn’t plan for all 40 users to need access at the same time and it’s certainly not a fair assumption based off the information given in the question. Poorly designed and worded question that should be removed.
The real solution here would be to use 8 x 62 host subnets would require borrowing from the 3rd octet. I do not see another way to understand this.
I don’t know the boss of this company and based off the info presented in the question, wouldn’t automatically assume it’s okay to not support all 40 hosts simultaneously.
Where specifically in the Cisco supported CCNA material does it explicitly state that one can assume “approximately 30-40 users” could also mean “only 30 users?”
There is just no chance I could return to my boss and tell him that we can only support “30-40” simultaneous internet connections and that be acceptable. No. The only solution still is to borrow more host bits and use a larger subnet. The question does not state that we can’t use larger subnets, but a correct answer is not provided to choose from.
E. None of the above