CCNA – Subnetting
Here you will find answers to CCNA Subnetting Questions
Note: If you are not sure about subnetting, please read my Subnetting tutorial.
Question 1
Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87
E – 201.45.116.159
F – 217.63.12.192
Answer: B C D
Explanation
A subnet mask of 255.255.255.224 has an increment of 32 (the binary form of the last octet is 1110 0000) so we can’t use numbers which are the multiples of 32 because they are sub-network addresses. Besides, we can’t use broadcast addresses of these sub-networks (the broadcast address of the previous subnet is calculated by subtracting 1 from the network address). For example the network address of the 2nd subnet is x.x.x.32 then the broadcast address of the 1st subnet is 32 – 1 = 31 (means x.x.x.31).
By this method we can calculate the unusable addresses, which are (notice that these are the 4th octets of the IP addresses only):
+ Network addresses: 0, 32, 64, 96, 128, 160, 192, 224.
+ Broadcast addresses: 31, 63, 95, 127,159, 191, 223.
Question 2
Which of the following host addresses are members of networks that can be routed across the public Internet? (Choose three)
A – 10.172.13.65
B – 172.16.223.125
C – 172.64.12.29
D – 192.168.23.252
E – 198.234.12.95
F – 212.193.48.254
Answer: C E F
Explanation
Addresses that can be routed accross the public Internet are called public IP addresses. These addresses belong to class A, B or C only and are not private addresses.
Note:
Private class A IP addresses: 10.0.0.0 to 10.255.255.255
Private class B IP addresses: 172.16.0.0 to 172.31.255.255
Private class C IP addresses: 192.168.0.0 to 192.168.255.255
Class D addresses are reserved for IP multicast addresses and can’t be routed across the Internet (their addresses begin with 224.0.0.0 address).
Also we can’t use 127.x.x.x address because the number 127 is reserved for loopback and is used for internal testing on the local machine.
Question 3
A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two)
A – 255.255.255.0
B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
E – 255.255.255.192
F – 255.255.248.0
Answer: B E
Explanation
We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask (29=512), leaving 7 bits for hosts which is 27-2 = 126 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.
We can increase the sub-networks to 1024 ( 1024 = 210), leaving 6 bits for hosts that is 26= 64 > 50 hosts. This scheme satisfies the requirement, too -> E is correct.
Notice: The question asks “The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet” but this is a typo, you should understand it as “”The company needs a minimum of 300 sub-networks and a minimum of 50 host addresses per subnet”.
Question 4
Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)
A – 115.64.8.32
B – 115.64.7.64
C – 115.64.6.255
D – 115.64.3.255
E – 115.64.5.128
F – 115.64.12.128
Answer: B C E
Explanation
CIDR stands for Classless In4ter-Domain Routing, the difference between CIDR and VLSM is slim and those terms are interchangeable at CCNA level.
To specify which IP addresses fall into the CIDR block of 115.64.4.0/22 we need to write this IP address and its subnet mask in binary form, but we only care 3rd octet of this address because its subnet mask is /22.
(x means “don’t care”)
Next, we have to write the 3rd octets of the above answers in binary form to specify which numbers have the same “prefixes” with 4.
4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100
We can see only 7, 6 and 5 have the same “prefixes” with 4 so B C E are the correct answers.
Question 5
Refer to the diagram. All hosts have connectivity with one another. Which statements describe the addressing scheme that is in use in the network? (Choose three)
A – The subnet mask in use is 255.255.255.192.
B – The subnet mask in use is 255.255.255.128.
C – The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D – The IP address 172.16.1.205 can be assigned to hosts in VLAN1.
E – The LAN interface of the router is configured with one IP address.
F – The LAN interface of the router is configured with multiple IP addresses.
Answer: B C F
Explanation
VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.
By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) and can be assigned to hosts in VLAN 1.
For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into multiple sub-interfaces with multiple IP addresses -> F is correct.
Question 6
The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet if the ip subnet-zero command is configured on the router? (Choose three)
A – 172.25.78.243
B – 172.25.98.16
C – 172.25.72.0
D – 172.25.94.255
E – 172.25.96.17
F. 172.25.100.16
Answer: A C D
Explanation
If the “ip subnet-zero” command is configured then the first subnet is 172.25.0.0. Otherwise the first subnet will be 172.25.32.0 (we will learn how to get 32 below).
The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000).
First subnet: 172.25.0.0/19
Second subnet: 172.25.32.0/19
Third subnet: 172.25.64.0/19
4th subnet: 172.25.96.0/19
5th subnet: 172.25.128.0/19
6th subnet: 172.25.160.0/19
7th subnet: 172.25.192.0/19
8th subnet: 172.25.224.0/19
In fact, we only need to specify the third subnet as the question requested. The third subnet ranges from 172.25.64.0/19 to 172.25.95.255/19 so A C D are the correct answers.
Question 7
Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from router A?
A. 172.16.0.0/16
B. 172.16.0.0/20
C. 172.16.0.0/24
D. 172.32.0.0/16
E. 172.32.0.0/17
F. 172.64.0.0/16
Answer: A
Explanation
Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.
All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.
The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.
-> Only answer A has these 2 conditions -> A is correct.
Please do the necessary change in the solution for Question No 3 , The formula used for calculating the number of host is 2^k (where k is the number of 0 bits left in subnet mask).
That should be 2^k -2.
Pl do the needful and update if I am wrong.
@Mihir Keshav: in fact 128 is much bigger than 50 hosts so I ignored the (subtract 2). But I updated it so that no one will be confused in the future. Thanks for your detection!
The router will take route 10.1.3.3 to send packets with IP 10.1.5.65 Because you have to look for ip nearest to ip 10 .1. 5. 65 with highest subnetmask which is 10.1.5.64/29
Thanks for the material i have successfully completed CCNA Thanx a lot…….
It was very helpful
Hi Dear all,
Please solve questions no.5, 6 and 7 i am totally confused in these questions
Thanks in advance
Can some one verify this . Question is
Ip subnet zero is not configured on router. What would the I address of Ethernet 0/0 using the first available address from the sixth subnet of the network 192.168.8.0/29 ?
A. 192.168.8.25
B. 192.168.8.41
C. 192.168.8.49
D. 192.168.8.113
My answer is B 192.168.8.41 because the sixth subnet is 192.168.8.40 – 192.168.8.47 but their answer is C 192.168.8.49 which would be in the seventh subnet 192.168.8.48 – 192.168.8.55. Am I missing something here?
Guys, can some one helping me with this question ?
Given IP address 172.16.128/27
Find ip address for 50th host ?
Thanks Guys
Hi Raj,
0 subnet is not configured,
That’s why sixth subnet is .48 ,
so fst add. will be 192.168.8.49
(0)is not applicable according to the question.
so it is 8, 16, 24, 32, 40, 48, annd so on.
think, it is clear now.
good-luck.
Hi kid,
where is another octet ?
Put it and you will get the ans.
Only thing im confused on is summarization. Sometimes they break it down by how many networks to be summarized then other times they break down the common bits used. Can someone tell me the difference or send me a good website that breaks it down? Thanks in advance.
@LOOK
Which octet ? Do u mean the question is still lack of information. Coz that’s all my teacher gave me..??
Hi Kid,
Please verify it again.
Because, Ip v4 Add. have 32 bits.
You have only 24 bites. Where is another 8 bites ?
And /27 means
’1′ ——27 bites and ’0′ ——5 bites.
That’s why i am looking another 8 bites.
Otherwise how you will formulate the ip. address ?
Please, Let me know your verification. Thank-you.
Thanks to 9tut bcs I was studying Dumps n stuck at subnetting questions..
Here i got solution..
Wonderful….
Hi kid,
It may be 172.16.128.0/17
Not 172.16.128/27
If it is, let me know. Good-luck.
given that host a and host b are in different networks. When host A is communicating with host B, which step will host A take first?
A. send a TCP SYN and wait for the SYN ACK with the ip address of host b.
B.drop the data
C.create an ARP request to get a mac address for host B.
D.send the data frames to the default gateway.
which is correct ans pls send ans to this ID rabinson.r@laserwords.co.in
ans is D
please i would like a review on q4 and q6 simply dont understand your explanation i would want a clearer view or opinion for q4
@landon
Q4
115.64.4.0/22 (255.255.252.0)
is
115.64.4.0 ~ 115.64.7.255
which of the given options refer to IPs that are part of this range? options B, C and E.
Q6
172.25.0.0 divided into 8 subnets. there’s nothing said about subnet-zero so we assume it’s enabled (as it is by default nowadays).
we’re interested in finding out which one is the 3rd subnet.
let’s find out the subnet mask:
172.25.0.0 <= class B address, 16 bits are 1 by default.
divided into 8 (2^3) subnets <= 3 bits are 1 by default.
the subnet mask is /19 or 255.255.224.0
the increment is 32.0
first subnet is 172.25.0.0 ~ 172.25.31.255
second is 172.25.32.0 ~ 172.25.63.255
third is 172.25.64.0 ~ 172.25.95.255
we're interested in the third one.
which of the options are part of that subnet? A, C and D.
please read 9tut's subnetting tutorial and please make sure you understand perfectly the very last part (how to subnet fast).
http://www.9tut.com/subnetting-tutorial
Hi guys I need your help with the dumps please you can send it at pmahhwayi@webmail.co.za
I am trying to write my ccna via ICND1 and 2, I currently install equipment and configure, but am struggling to pass exams as I seem to somehow not read questions properly, my main downfall is accesslists and ip addressing, can anyone assist with sites to get relevant training material to assist me?
@xallax
1 octet for subnets: 2^8 = 128 subnets
isn’t it 256…
@jay
hey
2^8 is 256, yes. where did i type it wrong?
I dont understand no SHit ….. Im crazy right now ! tryin to understand this crap !
passed ccna today with 933 marks 9 tut works :)
@XALLAX plz xpln me Q5…for 114 host i need minimum of 128 host…thats satisfied by 2^7…
so additional seven bit i’ve to make as network bit…..then why they are saying..
“so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.”
@snap
because the class B address takes 16 bits by default (xxx.xxx.0.0)
now the subnet takes 7 bits, that leaves 32-16-7 bits for subnetting, 16-7, 9 bits
255.255.255. 1 bit (128)
to xalax and 9tut..
Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87
E – 201.45.116.159
F – 217.63.12.192
why D is an answer? i believe its option E.
hope for your reply.
Thanks.
sorry its options B and C only…
@jay
subnet: 255.255.255.224
increment: 32
ranges: 0~31, 32~63, … N*32 ~ (N+1)*32-1 … 224~255
A: is the broadcast address on that subnet
B: is a regular IP on the 64~95 subnet
C: regular IP on 32~63
D: regular IP on 64~95
E: broadcast address on 128~159. 159 is (N+1)*32-1
F: network address on 192~223. 192 is N*32
yup… your right… thanks.
by tutorial I learnt 2+2-1=?
and in question I got 179^8z-654y+x=?
:xD
Regarding Sub-netting..
The network 172.16.0.0/27 has been divided into equal subnets. The following IP addresse 172.16.1.64/27 comes under which subnet if the ip subnet-zero command is configured on the router?
in the above we are subnetting on both the octets. we have 4096 networks with 30 hosts.
How to know to which subnet 172.16.1.64/27 comes under?
@Shyamsrule…
CIDR /27 is in the fourth octet ( 27-23= 3 bits implying 224 , if 256-224= 32 block size )
Subnets: 0, 32, 64, 96, 128, 160, 192, …. Remember from one subnet to another we one network and one broadcast, so network 0 has broadcast network 31, with range of networks from 1-30 that can be assigned. You can not used the broadcast for any other purpose. It is true for even anothe block size like 32 as network, 63 as broadcast and range 33-62 subnets can be assigned to the network.
The issue with the subnetting concept is that you need to do alot of practice untill the materials sink into your head and you can quickly and logically identify them. It took me time to get them but I finally achieved it. Hope my explaination helps!
@Shyamsrule.. correction in typo … (27-24=3…)
@Johny
In refference to Question no 6 above, only one octet is subnetted.
But what if it is in the case of the following: 172.16.1.66/27 ?
Here we are subnetting both the octets (3rd and 4th octet)
In the 3rd octet block size will be 1, so we will have 256Networks
In the 4th octet block size will be 32, so we have 8networks
So totally we have 256*8=2048 networks. Right…
Now 172.16.1.66/27 belongs to which subnet ?
@Shyamsrule
I think 172.16.1.66/27 belongs to subnet 172.16.1.64 because, it falls in the range of 65-94 in the block size of 64. We are looking at the 4th octet bse our mask is /27 = 255.255.255.224.
@xallax. hi may ask if you have a latest dump..i will appreciate if you can send it to my email. i used to download it from examcollection but all i got is the trial version which is only have 5 questions..thanks a lot..
kits_ace13@yahoo.com
@anonymous
there is a topic on certprepare.com regarding your exact problem. please visit:
http://www.certprepare.com/forum/index.php?showtopic=1664
check mail
have a nice day
let me share with you more subnetting questions. this was done by our teacher in cbtnuggests http://www.informit.com/articles/article.aspx?p=1186094
client 1, client 3 and client 4 can communicate with server b and server a.client 2 and client 5 cannot communicate with either server.yesterday there was no problem with this communication.you use the ping utlity as seen beloe:
c:\>ping 172.17.20.35
pinging 172.17.20.35 with 32 bytes of data:
destination host unreachable.
destination host unreachable.
destination host unreachable.
Explain what could be the most possible cause of the program?
Hi all,
I am going to appear for the exam in next month,
Could anyone send me the latest dumps to my email ID – syed.abitheen@gmail.com
Thanks in advance,
Syed
Hi EVERYONE – I WOULD LOVE TO GET THE LATEST DUMP FROM ANYONE WHO CAN GIVE IT TO ME> Please!!! I am taking the exam on May 15th and really would appreciate it.
Thank you!!!!
erica3025@gmail.com
Erica
The default sebnut mask for class B address is: 255.255.0.0, now we have changed it to 255.255.252.0, if put both in binary they would be:255.255.0.0 ====> 1111 1111 . 1111 1111 . 0000 0000 . 0000 0000255.255.252.0 ==> 1111 1111 . 1111 1111 . 1111 1100 . 0000 0000 we have borrowed 6 bits to form the new sebnuts.
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@Robert
you can download them for free from this site. All the best in your exam. mine is soon too
thanks 9tut!=D
Hi all,
I am going to appear for the exam in this week,
Could anyone send me the latest dumps to my email ID – ashraf.pantho@yahoo.com
Hi All.
Regarding Question 7, the method used was:
*) Look for networks beginning 172.16 / possible answer A, B or C
*) Look for a smaller or equal subnet mask than the routes being summarised / which leaves answer A.
The answer is more a process of elimination than what I was learning.
In Todd Lammle’s CCNA book, he tells us to look for the block number, then minus from 256 to reveal the mask and from the mask we can work out the new subnet mask – but I can’t seem to get it to work for this example!
Please some one clear this up for me – I just don’t get it!
@IceBlue
was also thinking the explanation is quite vague
Here’s my take: considering the network range to be advertised, btwn 32-164 in 3rd octet, the corresponding block size is 256, thus the summary mask of /16