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
This is probably obvious but, what makes 10.2.1.3 a broadcast address? In question one.
Disregard
Hey Bill,
by solving que you will get 10.2.1.3 as broadcast address.As it is /30 it is point to point connection. SO it has only 2 valid host. we can see here N/w id is 10.2.1.0 ,First vaslid host is 10.0.1.1 ,Last valid host 10.2.1.0 , BroadCast address 10.2.1.3 …… we cant use Broadcast address as IP.
Hey Rutvik, I think you meant to write “Network address is 10.2.1.0, First valid host is 10.2.1.1 and last valid host is 10.2.1.2 which makes broadcast IP 10.2.1.3 which is not assignable to hosts.
Q1) 10.2.1.3 – for this /30, there are only 2 usable IP’s right? 2^2 =4 so increments of 4, 10.2.1.0 would be the network address, but 10.2.1.4 should be the broadcast right? SO 10.2.1.2 and 10.2.1.3 should be usable?
Good Question NI. Here is where you are mistaken. 2^2 is 4. That is correct. Which gives the /30 subnet 4 addresses. The addresses are in this case 10.2.1.0, 10.2.1.1, 10.2.1.2, and 10.2.1.3. You were forgetting the typical .0 Network ID. This leaves 10.2.1.1 and 10.2.1.2 as the usable IPs; 10.2.1.3 as broadcast and 10.2.1.4 as the network Id in the next /30 subnet.
I hope Q9 is not so vague in the exam!
“30-40 users per floor”
I calculated for 40 users per floor and got a different answer *rolleyes*
Question 9
i think answer must be ip address 192.168.0.0 255.255.255.192 32-27=6
64 (=2 ^ 6) hosts 64-2 =62(in fact 62 hosts)
Question #9 is actualy not solveable, if we do expect to only have permission to use the original 192.168.0.0/24 net. This is due to the fact, that we expect 30-40 host on 8 floors. Even if we try to diverde this subnet into 8 subnets, we do run into the issue of an address shortage.
8 Subnets eq 16 addresses blocked for NW-address + bc-address
8*30 users = 240 adresses
Which summs up to 256 addresses. This means that per floor only 30 ppl max can have an ip, which renderes the rest (30-40) unable to participate in the network. This means we need to get rid of the idea, that we only hold administartion over 192.168.0.0/24 but rather 192.168.0.0/x, wich would mean that answere D would be right if we asume we would be able to make use of 192.168.0.0/22
Q9 the answer is B which is ip address 192.168.0.0 255.255.254.0
Shouldn’t Question 9 be B? probably an error?
please send me question ccna 200-301
Q9
Answer is B
Q9-The question is 8 floors x maximum users is 40. Per floor you will need at least a /26 which gives 60 hosts per floor, if we multiply by 8 x 40 = 320 hosts. That means that we will need a total of /23 which is going to give us 508 hosts. So, I don’t agree with the answer of C, this is why I need Cisco’s answer and not my own by my guess is either D or B, I would go with D.
Let me know if I’m correct?
Q9-8 floors with a max of 40 per floor that is per floor you will need a /26 which give us 60 hosts, or if we are going to add all 8 floors that would be 8 x 40 = 320, for this we will need a /23 to give us 508 hosts. Answer C only give us /27 = 30 hosts max and does not provide sufficient IPs for each floor if max is 40. So, if we go with the same logic as Q6 the answer should be B, but again, closes to C is D. but not C. This is where I need to right answer according to stupid Cisco and not according to brilliant me.
Q9 -> Answer is B ?!
@9tut, is answer B?^^^^
@9tut
Please answer if Q9 is correct….we are paying premium.
@All: We still believe answer C is the best choice here. The questions said ” approximately 30-40 users per floor” so we can choose between 30 and 40 hosts.
@ All, unable to see the question and option of the answers.
Have to install any application to view questions. I can see only explanations.
Please guide me. Thanks
I Got the link of PDF. Ignore.
Thanks
Q9
@9tut your calculation in explanation is wrong. Why? Glad that you are asking:
1. For example, on 6 floors could be 37-40 users and on other 2 could be 30-36. Average number of users per floor would be 33+, 8*33=254 (at least in my example). You guys took that there are 30 users on every floor, whereas it was stated “approximately 30-40 users per floor” and by that it will be at least 31 users on at least one floor. With that said, there is in total, at least 241 users. With this, your segmentation can not “cover” all users.
2. Again, it is said “approximately 30-40 users per floor”. “Users”, in our case, means people if I am not mistaken. We are assuming that every user will have just one device (laptop, desktop), hopefully best case scenario. Your calculation hasn’t taken into account IP address for the gateway for every of your subnets. I assume that in IT terminology GW is not considered user. Because of that, to the number of users per floor you have to add one number to the assigned IP addresses which is GW’s IP.
3. If I am not mistaken, you haven’t reserved IPs for network and broadcast addresses. The way you calculated and segmented you would have to subtract 3 addresses (including GW addr mentioned before) from available pool of addresses for every subnet. With that said it means that 32-3=29 which means that only 29 addresses per subnet are available to, at least, 30 users per floor and that means that at least 1 user per floor will not be connected to the network.
It is said 8 floors, not 8 subnets. We are assuming that 8 floors means 8 subnets. /25 gives us 2 subnets with 127 IP addresses each. When we subtract 3 IPs (for network, broadcast and GW) we get 124 available IP addresses which can be assigned to user’s devices. With that said, 2 subnets multiplied with 124 users/subnet gives us total 248 users. When we divide 248 with 8, which is number of floors, we get 31 users per floor and that falls under statement “approximately 30-40 users per floor”. Because of this correct answer is D.
However, we do not have one crucial information and that is total number of users which could span between 8*30=240 and 8*40=320. D is correct answer only if total number of users doesn’t exceed 248.
Unfortunately for our heads, question is pretty vague.
Cheers!!!
P.S. We can additionally segment subnets by using vlans where one vlan is assigned to one floor.
We reviewed Q9 again but still see C is the best choice. We updated the explanation a bit to explain why answer D is not correct. Although with answer C, we cannot cover 40 users/floor but we can cover 30 users/floor and this is the best choice.
I agree with C being the answer for Q9. Here is why:
192.168.0.0 is a class C network. So the mask is 255.255.255.0
If we subnet it then we’ll borrow from the last octet. For 8 networks, we’ll borrow 3 bits. That leaves us the 5 bits for hosts which will give us around 30 hosts. This meet the requirement of the question.
Question 6. Don’t understand, Hope can be elaborated more.
The Number of users is between 24 and 29, that means we will use /27 that give maximum of 30 users,
/25 is the summary of 4 subnetwork of /27
Q9, C is correct
My bad, I was doing calculations and have to say Q9 B is correct
Hello,
I don’t find the question on this section : IP Address & Subnetting Questions
I can only find the explanation.
Can you fix it if possible please ?
Thank you in advance.
192.168.0.0 255.255.255.224 8 subnet x 30 hosts 240 hosts
192.168.0.0 255.255.255.128 2 subnet x 126hosts 252 hosts
8 floors x40 users 280 users.
It is right?