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.
Can any one give me answer and explanation for following question ?
(plz mail on: sagheer3001@yahoo.com)
The ip Subnet Zero command is not configured on a router . What would be the Ip address of Ethernet0/0 using first available address from the SIXTH subnet of the network 192.168.8.0/29
A 192.268.8.25
B 192.268.8.41
C 192.168.8.49
D 192.168.8.113
Can any one give me answer and explanation for following question? Which OPTION IS TRUE,
plz mail me on: sagheer3001@yahoo.com
The ip Subnet Zero command is not configured on a router . What would be the Ip address of Ethernet0/0 using first available address from the SIXTH subnet of the network 192.168.8.0/29
A 192.268.8.25
B 192.268.8.41
C 192.168.8.49
D 192.168.8.113
Here is the answer
– 192.168.8.41
Method:- You are aware of the fact that /29 is a way to represent the subnet mask that is the bit notation. So Buddy this implies he subnet mask is 255.255.255.248
When you get a question like this Try to get the increment & get the network Ranges
(a) within the 4th octet of the subnet mask you can see that ==> 11111000
^
^
^
^
So this is the increment ( Increment =8)
So then we can find the Network Ranges
(b) Network Ranges
192.168.8.0 —192.168.8.7
192.168.8.8 —192.168.8.15
192.168.8.16 —192.168.8.23
192.168.8.24 —192.168.8.31
192.168.8.32 —192.168.8.39
192.168.8.40 —192.168.8.47
Final one is the 6 th network range. we can’t use 192.168.8.40 because it is the network id & 192.168.8.47 because it is the broadcast ip
So the first usable ip of the 6th network range is 192.168.8.41
I am a srilankan guess you are an indian anyway buddy here is the answer method to do it
Here is the answer
– 192.168.8.41
Method:- You are aware of the fact that /29 is a way to represent the subnet mask that is the bit notation. So Buddy this implies he subnet mask is 255.255.255.248
When you get a question like this Try to get the increment & get the network Ranges
(a) within the 4th octet of the subnet mask you can see that ==>
1 1 1 1 1 0 0 0
(32) (16) (8) (4) (2) (1)
Increment is 8
So then we can find the Network Ranges
(b) Network Ranges
192.168.8.0 —192.168.8.7
192.168.8.8 —192.168.8.15
192.168.8.16 —192.168.8.23
192.168.8.24 —192.168.8.31
192.168.8.32 —192.168.8.39
192.168.8.40 —192.168.8.47
Final one is the 6 th network range. we can’t use 192.168.8.40 because it is the network id & 192.168.8.47 because it is the broadcast ip
So the first usable ip of the 6th network range is 192.168.8.41
hi shaneera.what about the part of the question that says “The ip Subnet Zero command is not configured on a router”?
So my guess would be, subnet 192.168.8.8-192.168.8.15 is our first and 192.168.8.48-192.168.8.56 our last making 192.168.8.49 our first usable IP address in the SIXTH subnet.
Cheers.
hi guys, im sitting in this coming week, can anyone share the latest dump?
Dova Dude U R Ryt . I was Just Concentrating about the sub netting part only. So U r correct Here Thank z Buddy
Please explain me this question i dont know that how to understand this question?
Which two addresses can be assigned to a host with a subnet mask of 255.255.255.0?(choose two)
A:113.10.4.0
B:186.54.3.0
C:175.33.3.255
D:26.35.2.255
E:17.35.36.0
Answer is B & D
Hello, can someone please help me on Q6:
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
I understand the 32 increment but why is D correct? I would thought that 172.25.94.255 is a broadcast address, as i read in other topics on 9tut, even if we only care about the 3rd octet, if the last octet is 255 then its a broadcast address. Please help.
For Q3 why not C?when the company needs at least 300 subnets? Thanks.
sorry but can you guys help me with Q5 too? why is it 172.16.1.25 can be assigned to hosts in VLAN1?correct me if i’m wrong but isn’t it vlan2 the first subnet cause it has 114 hosts? thanks again
Hi, nobody really explain the summarization. there is one example which I dont get it? Could anybody explain to me?
3 routers: 10. 0.0.0, 10.0.2.0, 10.0.3.0 why the anser is 10.0.0.0./22??? (22- 254)
Thank you
@ KK: The explanation is as below:
10.0.0.0 = 00001010.00000000.00000000.00000000
10.0.2.0 = 00001010.00000000.00000010.00000000
10.0.3.0 = 00001010.00000000.00000011.00000000
It ‘s obvious that the first 00001010.00000000.000000 is common so 8 + 8 + 6 = 22
I hope it has been cleared for you.
@ Yoursef
I got binary but why 22, sorry still didnt get it. :-(
Does first octet mean 8 but wnat about rest 8 and 6?
Could you please explain that as well? Why did you put only 8+8+6 and not add the other 8? Sorry
Thank you
@Yourself,
I got it now, 8(same binary)+ same+ only 6 are same. thank you
thank in millions to u all for ur contributions. cheers
Just looking at that question three above. You could get the maths all correct and the answer wrong by interpreting the question incorrectly. Too many of the questions in the CCNA are worded vaguely :(
Q1. how many host are possible with a class c address and subnet 255.255.255.252?
ans is : 124
Q2. which network mask should you place on a class c address to accommodate a user requirment of two subnets with a maximum of 35 host on each network?
ans : 255.255.255.192
kindly explain
cn any1 plzz clear my dbt asap.. i nd it plzz… thank u
@teddy
Q1: the subnet mask 255.255.255.252 will provide only 2 hosts.( n=2 , 2^2 =4 , and 4-2=2)
Q2: User required a maximum of 35 hosts, hence a mask of /26 will provide up to 64 hosts for each subnet. that means 255.255.255.192
@zahir – thank u soo much ….. god bless u….
To Shiela
how many hosts and submasks in 172.16.0.0/23 subnetted /27options are?
16 subnets and 30 host!
How did you get that?
================
This is the easy way to you to compute the subnets;
given: /27 & /23
just subtract /27 – /23 = 4 (this will be your increment)
then use the formula 2^n, n= increment
2^4 = 16 sub
and for host obviously /27 have 30 usable IP’s, therefore the answer is;
16 sub & 30 hosts
CHEERS!!!
Here is a table I use for subnetting:
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
/25 /26 /27 /28 /29 /30 /31 /32
Example: /26 is 255.255.255.192 increments of 64 subnets are 0 64 128 etc….
/26 all the hosts should fall within the 0 64 128 etc… subnets, except for the actual subnet and broadcast address. This should quickly tell you if the hosts are in the same subnets or not based on subnet mask.
I would like to find route summary address but iam not it what is the process
Given the subnet mask of 255.255.224,what would be the following addresses can be assigned to network host (choose three)
a. 15.234.118.63
b. 92.11.178.93
c. 134.178.18.56
d. 192.168.16.87
e. 220.45.116.59
f . 2.1.63.12.192
this is a latest cert king dump question
the answers are b.c.d
how are the answers BCD
I just added your blog site to my blogroll, I pray you would give some thought to doing the same.
PLEASE EXPLAIN THIS QUESTION. THANKS
The network administrator is asked to configure 113 point-to-point links.Which IP addressing scheme
best defines the address range and subnet mask that meet the requirement and waste the fewest subnet
and host addresses?
A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252
@PLEASE EXPLAIN THIS QUESTION
I suggest u use D.
You need 113 subnets with 2 networkhosts (/30), so /23 gives you 512 hosts.
Divided by 4 (255.255.255.252), gives you 128 subnets, with blocks of 4, and 2 usable host addresses. Perfect for PtP.
Hi 9tut, I just want to clarify this question:
Address Range: 10.188.31.0/23
You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each. which IP address range
meets these rquirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
> the answer was B
The answer should be /27 right?
/25 = Increment of 128,
/26 = increment of 64,
/27 = increment of 32.
and the question was as many as 30 hosts each subnet.
> I just want to double check with you.
Thanks
@Hmmm: Yes, A is correct as we only need 5 bit 0 -> /27.
hallo everone,am about to sit for the ccna composite exam,am in tanzania east africa,i am kindly asking anyone and everybody for your help that i get the latest ccna dump that is valid.Here in Africa we dont have a wide spread use of credit cards to pay online for the dumps,so its very tricky downloading from sites,coz they require a sort of payment or so,and i cant afford to fail on the first sitting,370$ is the fee here,may you please help me all of you,send me all the relevant dumps to chriscathbert@yahoo.com,thank you very much for sharing all this info,its great of all you members,thanks once more
i hv a comment on Q5…
option F – The LAN interface of the router is configured with multiple IP addresses.
i think it got a typo mistake, it need to say “…..is configured with multiple IP addresses SUB-INTERFACES” cuz sub-interface is different than multiple IP addresses, we can use more than one IP address at the same interface by using the ip address xxxxxxx secondary command, n its totally differnet than sub-interface.
plz correct me if im wrong
cheers
You are the network administrator at CiscoTests. CiscoTests has been provided with the network address 165.100.27.0/24. The CiscoTests CEO wants to know how many subnetworks this address provides, and how many hosts can be supported on each subnet.
A. One network with 254 hosts.
B. 254 networks with 254 hosts per network.
C. 65,534 networks with 255 hosts per network.
D. 30 networks with 64 hosts per network.
E. 254 networks with 65,534 per network.
Any one? Please give me a explanation. I think its B.
255.255.255.0
what is the subnet mask and network
394.
You are working in a data center environment and are assigned the address range 10.188.31.0/23.You are
asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30
hosts each.Which IP address range meets these requirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
Answer: B
———
y its not a???
pls help me wit dis ques!
Which 3 ip addres can be assignd to hosts if the subnet mask is /27 and subnet zero is usable? (choose 3).
A-10.15.32.17
b-17.15.66.128
c-66.55.128.1
d-135.1.64.34
e-129.33.192.192
f-192.168.5.63
ans-acd
i tink d ans shld be c,d or d ques is wrong.
Pls help hv xams 2mao
which of the fnlowin ip addres can be asignd to the host? (choose 2)
a-205.7.8.32/27
b-191.168.10.2/23
c-127.0.0.1
d-224.0.0.10
e-203.123.45.47/28
f-10.10.0.0/13
ans-bf why?
10.1.1.0/24
Is this considered as A class or C class ip address.
I know that it is subnetted but can somebody explain me how does the router identify that it is a Class A or Class C address.
thanks
Please note that question 3 (CCNA Subnetting) has an error.
if the answers are to be B,E then the question should say minimum 50 hosts per subnet, not maximum 50 hosts
Peace!!!! from Bashment
You are working in a data center environment and are assigned the address range 10.188.31.0/23.You are
asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30
hosts each.Which IP address range meets these requirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
Answer: B
please help me with this i think the answer is A
@dinesh
the answer should be “A”
@Bashmenty2k
B and E are correct. Why?
if we take maximum 50 host E is correct but we have to select 2 answers the next subnet is B.
according to u if it says minimum 50 host but say if we need 100host, then E will be incorrect.
i don’t know if i explained it clearly, go thru the explaination u will understand.
@mo
10.1.1.0/24
Is this considered as A class or C class ip address.
I know that it is subnetted but can somebody explain me how does the router identify that it is a Class A or Class C address.
thanks
—————————————
I think it’s A class ip add. The class of a ip add is recognized by first octect and the first octect of A class is 10 (or 0xxxxxxx in binary). Addtional, /8 is just default subnet mask of A class ip add, so /24 is not importance.
Thx 9tut. I have been read this site every day. :D
I saw a strange question from a dump:
Workstation A has been assigned IP address 192.0.2.24/28 Workstation B has been assigned IP address 192.0.2.100/28 The 2 stations connect with straight thru cable. What 2 things must be changed for conectivity?
answer
Replace straight with crossover cable
Change subnet mask to /25
I dont get the change the mask to /25. .24 and .100 are both in the /28 range. Whats going on with this?
Thanks.
weird because I tried it out in the PT and it worked with a /25 but not a /28 am I missing something here?
because a /25 will break the subnets into only 2 and this is required for a PC?
I think I get it, they got to be in the same subnet to communicate..sorry :) I been pinging through routers all day so got off track
hi 9tut i have a question about subnetting :
if a ethernet port of a router was assigned an ip address 172.1.1.1/20. How many host will be available in this subnet?
I know the answer is 4094 but i want to know how do i get to this number
16*255=(4080-2)=4078 isnt answer ?