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CCNA Subnetting Questions ­ 22966 ­ The Cisco Learning Network
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CCNA Subnetting Questions
Created by Paul Stewart ­ CCIE Security on Jan 7, 2014 3:46 PM. Last modified by Paul Stewart ­ CCIE Security on Jan 19, 2014 5:06
AM.
Visibility: Open to anyone
1. Given the IP address of 172.16.1.1 with a mask of 255.255.255.0­­How many total subnets could be created? (assume all
subnets use the same subnet mask)
65536
254
256
64
2. Represent /26 in dotted decimal format.
255.0.0.0
255.255.255.192
255.255.255.0
255.255.255.128
3. What address Class does 172.16.33.1/24 belong to?
Class
Class
Class
Class
A
B
C
D
4. When calculating usable hosts per subnet, the following formula is used 2^bits ­ 2. For what reason is two subtracted?
(choose two)
Broadcast
Multicast
Unicast
Network
5. Your organization is designing a Wide Area Network. Locations have varying numbers of hosts. The largest network will
have no more than 55 hosts. What subnet mask accomplishes the goal and maximizes the number of subnets that may be
created?
255.255.255.192
/25
255.255.255.224
/27
6. How many hosts can be located on a network, where the IPv4 netmask is 27 bits?
27
30
32
5
7. What are two ways to represent a network mask that would allow 14 hosts?
/14
255.255.255.240
255.255.14.0
/28
8. How many hosts can be addressed on 10.0.0.0/16?
16
254
65536
65534
9. Convert the following binary to decimal­­01101101.
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225
109
1101
112
10. Choose the true statements (choose two).
Odd numbers have the least significant bit set to 0
Even numbers have the least significant bit set to 0
Odd numbers have the least significant bit set to 1
Even numbers have the least significant bit set to 1
11. What subnet mask will allow for 128 hosts on a subnet? (choose the best two answers)
/25
/24
255.255.255.0
255.255.255.128
12. Based on 1.1.1.0/24, the IP address would be:
Class
Class
Class
Class
A
B
C
D
Answers
1. Given the IP address of 172.16.1.1 with a mask of 255.255.255.0­­How many total subnets could be created? (assume all subnets use the same
subnet mask)
65536
254
256 <­­
64
2. Represent /26 in dotted decimal format.
255.0.0.0
255.255.255.192 <­­
255.255.255.0
255.255.255.128
3. What address Class does 172.16.33.1/24 belong to?
Class A
Class B <­­
Class C
Class D
4. When calculating usable hosts per subnet, the following formula is used 2^bits ­ 2. For what reason is two subtracted? (choose two)
Broadcast <­­
Multicast
Unicast
Network <­­
5. Your organization is designing a Wide Area Network. Locations have varying numbers of hosts. The largest network will have no more than 55
hosts. What subnet mask accomplishes the goal and maximizes the number of subnets that may be created?
255.255.255.192 <­­
/25
255.255.255.224
/27
6. How many hosts can be located on a network, where the IPv4 netmask is 27 bits?
27
30 <­­
32
5
7. What are two ways to represent a network mask that would allow 14 hosts?
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/14
255.255.255.240 <­­
255.255.14.0
/28 <­­
8. How many hosts can be addressed on 10.0.0.0/16?
16
254
65536
65534 <­­
9. Convert the following binary to decimal­­01101101.
225
109 <­­
1101
112
10. Choose the true statements (choose two).
Odd numbers have the least significant bit set to 0
Even numbers have the least significant bit set to 0 <­­
Odd numbers have the least significant bit set to 1 <­­
Even numbers have the least significant bit set to 1
11. What subnet mask will allow for 128 hosts on a subnet? (choose the best two answers)
/25
/24 <­­
255.255.255.0 <­­
255.255.255.128
12. Based on 1.1.1.0/24, the IP address would be:
Class A <­­
Class B
Class C
Class D
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Gus Jan 18, 2014 12:26 PM
Paul..I couldn't wait to complete this exercise.
So I left aside for a moment my studies of Port Security and came to do this exercise as a test for me.
I scored 9 0f 10 correct. I missed number one, I answered 254..your answer 256.
Great practice.
Thanks,
Gus
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Jerry Jan 18, 2014 12:41 PM
for number 7. /28 is also an answer you forgot an arrow there.
Got two wrong. Number 10. Can you explain what that is trying to say?
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Uma Shankar V. Jan 19, 2014 1:51 AM (in response to Jerry)
Hi Jerry,
For Q10, answer is, for even LSB set to 0 and 1 for odd numbers.
If you convert the decimal into binary 0 ­ 0000 0000 [with values of 128 64 32 16 8 4 2 1].
To get an odd number, LSB should set to 1 and for even number, LSB should set to 0.
Ex:
For Value 55: equivalent binary value is 0011 0111.
For Value 66: equivalent binary value is 0100 0010.
HTH
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sino Jul 15, 2016 5:38 AM (in response to Uma Shankar V.)
thanks men
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Paul Stewart ­ CCIE Security Jan 19, 2014 5:09 AM (in response to Jerry)
Jerry,
Thanks for pointing out #7. It should now correctly show both correct answers. Regarding #10, it is more about
binary than subnetting per se. Uma Shankar V. correctly answered your question. Any even number converted to
binary has a 0 as the least significant bit. Odd numbers have this set to 1. For example
0
2
4
6
0000
0000
0000
0000
00000
00010
00100
00110
1 0000 0001
3 0000 0011
5 0000 0101
Make sense?
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Jerry Jan 19, 2014 12:09 PM (in response to Paul Stewart ­ CCIE Security)
makes sense now
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Uma Shankar V. Jan 19, 2014 1:47 AM
Perfect. Scored 12 out of 12!!! One small correction, for Q7, two answers are correct [255.255.255.240 , /28], but only one
highlighted
Thank you, paul. Looking forward for more questions!!
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Paul Stewart ­ CCIE Security Jan 19, 2014 5:10 AM (in response to Uma Shankar V.)
Uma Shankar V,
Thanks. I have updated #7. I'll try to post more questions as time permits.
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Uma Shankar V. Jan 20, 2014 7:26 PM (in response to Paul Stewart ­ CCIE Security)
Thank you, sir
!!!
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ARandall Jan 21, 2014 8:30 AM
Thanks Paul....great subnetting exercise.......please post more opportunities and increase the level of difficulty.
Alvin
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Stefan Feb 17, 2014 8:59 AM
just started re­study for ccent & ccna
enjoyed the questions and only got 1 wrong #10
understand now tho..
Many Thanks
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James Feb 27, 2014 1:16 PM
Great question to see where i stood with this... SCORED 12 OUT OF 12... The only tricky one was 10... but i just had to
remember the lowest bit value is one...
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maninder negi Feb 28, 2014 1:45 PM
Hi Paul....
can you please help me getting the propers subnets for below question....
10.0.0.0/3
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Paul Stewart ­ CCIE Security Mar 9, 2014 1:07 PM (in response to maninder negi)
maninder negi,
That's a little tricky. A subnet would, by definition, be a subset of a classful address space. 10.x.x.x is a class A
address, so it would be a /8 when we look at it classfully. Since it is given as a /3, that means it is a super net of
multiple class A addresses.
To break this down.
10 in binary is 0000 1010
I have underlined the first three bits.
Given this is 10.0.0.0 /3, that means that 0 (0000 0000) through 31 (0001 1111) would be valid first octets. But then
again, 0 is never valid as a first octet. and 1­9 and 11­31 wouldn't be subnets of 10.x.x.x (because subnet assumes
we are a subset of a classful network). 1­9, 11­31 are other valid supernets in the same 10.0.0.0/3 range. So then
we look at 10.0.0.0 and its subnets. Then it would depend on the mask given.
So, I think that is a bad and/or incomplete question. Does that help?
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maninder negi Mar 26, 2014 3:47 PM
Hi Paul ,
Thanks for the reply....:)
So you mean that 10.0.0.0/3 can never be any subnet ???
Can you please let me know that below mentioned subnet is correct ??
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0.0.0.0/1
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paul stewart Mar 26, 2014 3:57 PM (in response to maninder negi)
Hi Pual ,
Ignore my last last reply....
Suppose i have host address as 10.0.0.1/3 so it is valid host or something is wrong .
can you please provide a example of host having subnet less then 8 ( like 10.0.0.1/3 ) with proper subnetting or it is
not possible ??
Can you please let me know that below mentioned subnet is correct ??
0.0.0.0/1
Many Thanks in advance ....
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chris Jun 18, 2014 8:24 AM
Please help with number 11. Maybe I'm over thinknig it. It's the only one I missed.
Thanks!
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Henry Chan Jan 16, 2017 6:34 PM (in response to chris)
actually 7 bit can be 126 host only ( need ­2 ) so the question need 128 hosts. u need 8 bit.
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chris Jun 18, 2014 8:48 AM
Disregard my question. I used the formula 2^­2 and came up with 126. I didn't thin about the 2 bits that are reserved.
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kamal Jul 14, 2014 11:15 AM
it is giving to me very good exercise
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raj Aug 26, 2014 2:13 AM
sir, why do we need an IP address for computers if we need only to communicate within a single subnet,as we know switch
work on MAC address.
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Craig Sep 5, 2014 10:22 AM
#11 got me, tricky....255.225.255.128 = 128 Host (BUT wait ­2 for network and broadcast) = 126 so the answer is
255.255.255.0
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Madvesha Sep 7, 2014 6:15 PM
Thank you very much Paul, Good excerise for me :)
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Alex Sep 11, 2014 5:19 AM
Thanks Paul, I am new to subnetting and this helped me out a lot. Cheers
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mohammed Nov 17, 2014 10:38 AM
12­9 (3 wrong)
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Iyas Jan 10, 2015 3:32 AM
1st i want to thank you (Paul) for these excerise, Am still smurf in CCNA
CCNA? and can you Help me For Q7 to Explane to me the answer.
, But i get SCORED 9 ­ 12 is it good for begginer
Thank you every 1
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L1onkin Jun 1, 2015 2:58 AM
Q5, should it not be /25? 55 hosts, using the formula 'The number of hosts per subnet = 2k – 2 (k is the number of bits “0″
in the subnet mask)', so 2k – 2=55, k=7, therefore there should be 7 bits 0s in the last octet hence mask should be /25?
Thanks
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L1onkin Jun 1, 2015 3:02 AM
Sorry, ignore my last post
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mai Sep 15, 2015 4:10 AM
Hello
Please I would like to ask
what is
12. Based on 1.1.1.0/24, the IP address would be:
why not class C?
Thanks
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GB Sep 17, 2015 10:10 AM (in response to mai)
Class A addresses are in the 0 to 126 range for the first octet. The mask for a class A address is /8. The problem
has /24 which can be ignored because the question is asking what class that address is in. a /24 meens this class A
address is subneted using 16 bits from the host.
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Cathbert Sep 18, 2015 3:45 PM
Thank you Paul and everyone participated on this, it really makes sense.
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Nelson Nov 3, 2015 3:47 AM
Thanks for this, I have 12 correct answers
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Mikyrazy Nov 6, 2015 6:26 AM
Thanks a lot for this... had 11 out of 12.. only qtn 8.... chose 3rd ans instead of 4th....
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navy Nov 24, 2015 11:58 AM
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#1 is a confusing question. It is asking about ip 172.16.1.1 with a mask of 255.255.255.0. So from the question, there is
only 1 subnet and within that subnet there is 254 valid hosts. So how can you get 256 subnets?
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Shweta.Bhati Nov 30, 2015 10:09 PM (in response to navy)
@navy
172.16.1.1 with a mask of 255.255.255.0
Class B Address that Means ­
N.N.H.H = 8.8.0.0
Network Bits Borrowed = 8 (8.8.8.0) (bits left for Hosts = 16­8=8 means 256­2=254)
8=256 Networks / Subnets with 254 Hosts
hope it helps...
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AlexMcN Sep 23, 2016 12:24 PM (in response to Shweta.Bhati )
Shweta.Bhati,
I'm sorry, you say a 255.255.255.0 mask is a class "B" network? I'm new to this but so far I've been thinking
that 255.255.0.0 is the mask for the class B. I'm obviously missing something.
So if I'm not totally wrong here, I agree with the people who say 254 possible hosts.
Thanks
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Eddie Jan 31, 2016 4:52 PM
Isn't #11 a /25 mask? a /24 mask will have 254 hosts
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CHARAN G Feb 12, 2016 11:38 AM (in response to Eddie)
they've asked 128 hosts on a subnet..if you choose /25 then we will get 128hosts but valid hosts are 128­2(network
& broadcast)=126...so we chose /24
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jemal mohammed Feb 29, 2016 12:45 AM
nice question. but i got 11 out of 12. but i want you show me the steps for the first question. thanks
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Ankit singh May 6, 2016 3:37 AM
Scored 12/12
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Lynn McAllister May 25, 2016 4:36 PM
I think I may have found a typo in question #5. it states the answer is 255.255.255.192 /25. I thought 255.255.255.192 = cidr /26 &
255.255.255.128 = cidr /25 . I am confused on this one, please let me know how to calculate this so that my answer matches what is
listed?
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Rushlan May 27, 2016 10:50 AM (in response to Lynn McAllister)
/25 is another option which is 255.255.255.128 , it has nothing to do with 255.255.255.192.
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abyrd Jun 1, 2016 1:44 PM
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Thanks for doing this. I'm still not understanding #1. If a network admin is handed 172.16.1.0 255.255.255.0, you can get
64 point­to­point networks out of that address space. I believe @shweta.bhati has the right answer, it just seems like the
/24 is clamping down the first 3 bytes.
Also, I swear I saw that RFC1918 addresses are
10.x.x.x ==> /8
172.16.x.x ==> /12
192.168.x.x ==> /16
Which makes things more interesting.
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sino Jul 15, 2016 5:40 AM (in response to abyrd)
yes i still dont get it either even the last question
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mohamed Jun 13, 2016 1:24 AM
good i like that
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orly Jun 14, 2016 6:17 AM
please i need more explanation for question 5 thank you
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Olela Jul 24, 2016 8:40 AM
Hi John ,
I scored 10 out of 12. This is a good practice. I missed no.'s 8 and 10
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Boz Aug 1, 2016 1:26 AM
A good number of questions. I am looking forward for more exercises. Many thanks.
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Aakash Budhiraja Aug 20, 2016 1:39 AM
5. Your organization is designing a Wide Area Network. Locations have varying numbers of hosts. The largest network will have no more than 55
hosts. What subnet mask accomplishes the goal and maximizes the number of subnets that may be created?
255.255.255.192 <­­
/25
255.255.255.224
/27
In this question, It is saying largest network will have no more than 55 hosts. But your answer is 255.255.255.192. If we calculate this the
answer will be 62 hosts per subnetwork.. So this wrong.. The correct answer is 255.255.255.224. If we calculate this the answer will be 30
hosts per subnetwork.
If I am wrong please let me explain this.
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Ivan Galindo Dec 2, 2016 5:01 PM (in response to Aakash Budhiraja)
I understand on the question #5 that maximum allowed host required on the largest network are 55. But in the
255.255.255.192 there are a maximum allowed hosts of 64. Which it means there are enough hosts to supply the largest network.
To me it is correct.
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piyush kumar Oct 3, 2016 12:38 AM
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right answer of question 1 is 254
calculation:
Address: 172.16.1.1 10101100.00010000.00000001 .00000001
Netmask: 255.255.255.0 = 24 11111111.11111111.11111111 .00000000
Wildcard: 0.0.0.255 00000000.00000000.00000000 .11111111
=>
Network: 172.16.1.0/24 10101100.00010000.00000001 .00000000 (Class B)
Broadcast: 172.16.1.255 10101100.00010000.00000001 .11111111
HostMin: 172.16.1.1 10101100.00010000.00000001 .00000001
HostMax: 172.16.1.254 10101100.00010000.00000001 .11111110
Hosts/Net: 254 (Private Internet)
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punith Jan 31, 2017 1:11 AM
wow !!!! 12/12 ..good questions ,thank you paul
please post some tough questions....
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Shubham Apr 10, 2017 9:12 PM
Nice exercise I have got 10 correct and 1 skip .
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