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Subnetting Questions

April 1st, 2011 in ICND1 Go to comments

Here you will find answers to subnetting questions in ICND 1 exam

Note: If you are not sure about subnetting, please read my Subnetting tutorial.

Question 1

Refer to the exhibit. The goal of this network design is to provide the most efficient use of IP address space in a network expansion. Each circle defines a network segment and the number of users required on that segment. An IP subnetwork number and default gateway address are shown for each segment.

What are three problems with the network design as shown? (Choose three)

IP_subnetworks

A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30.
B – Interface fa0/1 has an invalid IP address for the subnet on which it resides.
C – Interface fa0/2 has an invalid IP address for the subnet on which it resides.
D – Network 10.1.2.0/25 requires more user address space.
E – Network 10.1.3.128/25 requires more user address space.
F – The IP subnet 10.1.1.0/30 is invalid for a segment with a single server.

 

Answer: A B D

Explanation

Answer A should be “Interface fa0/3 has an invalid IP address for the subnet on which it resides” to be the correct answer. But there is no better solution (answers C E F are obviously incorrect) so we must choose answer A.

Question 2:

If an ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A – 1024
B – 2046
C – 4094
D – 4096
E – 8190

 

Answer: C

Question 3:

Refer to the exhibit. The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of 255.255.255.224. The routing protocol in use is RIP version 2. Which address could be assigned to the FastEthernet interface on RouterA?

subnetting_1

A – 192.168.1.31
B – 192.168.1.64
C – 192.168.1.127
D – 192.168.1.190
E – 192.168.1.192

 

Answer: D

Question 4:

Refer to the exhibit. HostA cannot ping HostB. Assuming routing is properly configured, what could be the cause of this problem?

subnetting_2

A – HostA is not on the same subnet as its default gateway.
B – The address of SwitchA is a subnet address.
C – The Fa0/0 interface on RouterA is on a subnet that can’t be used.
D – The serial interfaces of the routers are not on the same subnet.
E – The Fa0/0 interface on RouterB is using a broadcast address.

 

Answer: D

Comments
Comment pages
1 2 10
  1. Shaytan
    September 17th, 2012

    Sorry meant fa0/3 interface not fa0/1.

  2. Rango
    September 20th, 2012

    i will be taking my Icnd1 in few weeks time, can someone help me with latest dumps materials at noamnotjoking@yahoo.com . thanx in advance

  3. andy
    September 25th, 2012

    Please Explain
    Question1: What is the first valid host on the subnetwork that the node 172.27.1.160 255.255.254.0 belongs to?

    Answer: 172.27.0.1

    Question2: What valid host range is the IP address 172.24.251.211/21 a part of?

    Answer: 172.24.248.1 through to 172.24.255.254

  4. Subh
    September 26th, 2012

    @Andy
    Q1 The mask is 255.255.254.0 .. so we need to break Network in the 3rd Octate and the increment is 2.
    So first consider the Network 172.27.0.0 which we need to break in 3rd Octate with increment of 2.. so it follows:
    172.27.0.0 –
    172.27.2.0 –
    172.27.4.0 –
    .
    .
    and it goes on..
    Now if you complete each address range it comes like this:
    172.27.0.0 – 172.27.1.255
    add one bit to the last octate of the last address and the next range will be
    172.27.2.0 – 172.27.3.255 .. and it goes on.
    So the node 172.27.1.16 clearly lies in the range 172.27.0.0 – 172.27.1.255 since we cannot use the first address in the range the first valid IP will be 172.27.0.1.

    Q 2. Consider the classfull address 172.24.0.0 with mask /21 or 255.255.248.0
    So we break the Network on the 3 octate with increment of 8
    The ranges that comes out..
    172.24.0.0 –
    172.24.16.0 –
    172.24.24.0 –
    .
    .
    .
    172.24.248.0-

    So the range that we are getting here is

    172.24.0.0 – 172.24.15.255
    172.24.16.0 – 172.24.23.255
    .
    .
    172.24.248.0 – 172.24.255.255
    Not considering the fist and the last IPs in the range valid host range will be 172.24.248.1 through to 172.24.255.254.
    I hope this helps..

  5. Subh
    September 27th, 2012

    Correction Q 2:
    we break the Network on the 3rd octate with increment of 8
    So the ranges that comes out..
    172.24.0.0 –
    172.24.8.0 –
    172.24.16.0 –
    172.24.24.0 –
    .
    .
    .
    172.24.248.0 –
    So the range that we are getting here is
    172.24.0.0 – 172.24.7.255
    172.24.8.0 – 172.24.15.255
    172.24.16.0 – 172.24.23.255 .. and so on..
    Sorry I missed here and there in hurry.. but the logic is correct 🙂

  6. mike
    October 10th, 2012

    Actually, I agreed with shariq becuase in the diagram 10.1.3.0/30 network is not being assigned in any of the host interface. No overlapping happened here as 10.1.3.1 IP is not been assign.

  7. N
    November 3rd, 2012

    How can Q1, A be correct?
    The Fa0/3 ip adress 10.1.3.1 is ON the 10.1.3.0 network, thus it doesnt overlap it.
    It cant however be the gateway for 10.1.3.128 /25 network as it is a different network..
    B & D are correct so could the third answer be F? I mean.. its not invalid per se, still i dont think Cisco approves of that kind of subnetting

  8. daPhly
    November 17th, 2012

    Hi guys, im writing my exam on frid da 23rd..so far it has been been me and cbt nuggets training videos, 9tut, examcollections and more subnetting questions
    can someboduy help me on this one.

    1.How many hosts could you put on each subnet of 1.0.0.0 if you used a mask of 255.25.248.0? How many subnets would you have created?
    a) 2048 useable hosts. 8190 subnets
    b) 2048 useable hosts. 8192 subnets
    c) 2046 useable hosts. 8190 subnets
    d) 2048 useable hosts. 8192 subnets

    answer: D
    I do understand the first portion of useable hosts, but the subnets?? can somebody give clearity on how they got 8192 subnets

  9. xallax
    November 17th, 2012

    @daphly
    255.255.248.0 is in CIDR notation /21
    255 = 1111 1111 (8 bits)
    248 = 1111 1110 (5 bits)
    8+8+5 = 21 bits

    it is a class A address (1.0.0.0) so the first 8 bits can not be touched.
    there are 21-8 = 13 bits for the subnets.
    2^13 = 2^10 * 2^3 = 1024 * 8 = 8,192 subnets

    how many hosts per subnet?
    there are 32-21 = 11 bits for the hosts
    2^11 = 2^10 * 2 = 1024 * 2 = 2048
    keep in mind that the first and last IPs on any subnet are reserved.
    that leaves 2048 – 2 = 2046 hosts per subnet

    the correct answer is: 2046 hosts, 8192 subnets (none of the options given by you…)

  10. daPhly
    November 19th, 2012

    Thanx xallax, actually I made a typing error its supposed to be 2046 useable hosts

  11. xallax
    November 20th, 2012

    @daphly
    you must be able to do subnetting math as seen above in less than a minute (20-30 seconds max).
    i’d recommend you start reading 9tut’s subnetting tutorial:
    http://www.9tut.com/subnetting-tutorial

    cheers

  12. daPhly
    November 27th, 2012

    thanx xallax and everybody..passed with 850 now m offf to ICDN2 exam! Wooooop woooop

  13. Srinidhi
    December 9th, 2012

    Max,Thanks for the information. I feel that not all csmotuers deserve a /48. I work for a Co-location facility and we also have several DSL and ATM csmotuers. So I am looking into the implementation for our csmotuers; If we follow the RFCs and guidelines, then I honestly feel that giving a Co-location customer /48 is a waste of address space if the customer only requires a handful of subnets, it would make more sense to give the customer a /56. We also implement HSRP for csmotuers that have IPv4 connectivity, so when the time comes to implement IPv6 and HSRP (There is anycast), but for now, we are looking at all options and the best scalable method; So in the case with HSRP and IPv6, it does not make sense configuring a vlan interface with a /64. I see that maybe configuring the SVI with a /120 and then routing a /56 or a /48 to the csmotuers router from the /120. I have reviewed many RFCs and the discussion for subnets that are longer than a /64 and what is breaks and so fourth, so I am just trying to get an idea what others have implemented.Thanks for your time.Chris

  14. Dallas Naijan
    December 10th, 2012

    I am totally having some unfortunate luck with all of my Cisco attempts for 640-802 and then stepping down to doing the two part 640-822 and 640-166. On my last two failed attempts with the 802 exam, I failed both exams by 10 points which caused me to go hysterical. When I decided to go for the 640-822 test, I studied hard for another 3 weeks with all sorts of tools provided:
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    – ExamCollection
    – VCE
    – subnettingquestions.org
    and I still ended up failing for with a 775/1000. I was for sure that I passed the exam when I was killing all of my practice material. I definitely do not know what is going on and this is extremely disturbing. I keep seeing others with great success stories, and I’m trying to reach that pinnacle also to relay my experiences, but damn, I’m in a hole I can’t get out of. Can anyone out there feed me some of the latest dumps for the ICND1 exam. I would definitely appreciate this as I’m trying to be a family member in the Cisco world.
    I can be reached easily at cehiem@yahoo.com

  15. Gemma
    December 11th, 2012

    Roermond, 10 nov. 2010.LS,As a 76-year-old I must teach myself all the neisseacry lessons about the PC. To acquire knowledge for the amateur is difficult. It is there but quite often for a layman very difficult to understand and to put into practize.Your explanation is for me of an enormous value; I am very grateful,as I do understand your directions, AND IT LEADS TO SUCCESS.Thank you very much.D.A. Bokhorst.The Netherlands.VA:F [1.9.13_1145](from 0 votes)

  16. rama
    December 19th, 2012

    Assuming a subnet mask of 255.255.248.0, three of the following addresses are valid host addresses.
    Which are these addresses? (Choose three.)
    A. 172.16.9.0
    B. 172.16.8.0
    C. 172.16.31.0
    D. 172.16.20.0
    Answer: ACD
    but 31 is broadcast address
    please anyone can explain me

  17. Anonymous
    December 20th, 2012

    @rama
    given the mask of 255.255.480.0, thus your range is 8 (256-248=8).
    the network is 172.16.8.0
    range of addresses is 172.16.8.1 thru 172.16.255.254, thus any addresses in between are valid which is ACD.

  18. rama
    December 20th, 2012

    @Anonymous thanks yes its true..

  19. Desyjamz
    December 21st, 2012

    thnx alot pass my ICND1 wth 912 yesta

  20. Desyjamz
    December 21st, 2012

    thnx alot passd my ICND1 exam yesta wth 912

  21. Anonymous
    December 29th, 2012

    The question 4 is wrong! You can have 2 questions that are right!
    The A and D.

    I say this because Host A is not on the same subnet as is default gateway

    Proof
    (32 Network) 33-46 (47 Broadcast) – This is where Default Gateway Is
    (48 Network) 49-62 (63 Broadcast) – This is where host A Is

    So that have 2 right answer’s

  22. Q3 I find Interesting
    December 31st, 2012

    Anyone want to know the reasoning as to why Q3 is “D”, 192.168.1.190

    192.168.1.0/27
    255.255.255.224

    Increment is 32

    192.168.1.0 – 192.168.1.31
    192.168.1.32 – 192.168.1.63
    192.168.1.64 – 192.168.1.95
    192.168.1.96 – 192.168.1.127
    192.168.1.128 – 192.168.1.159
    192.168.1.160 – 192.168.1.191 – this being the network in question
    192.168.1.192 – 192.168.1.223

    The question is a mind warp, even mentioning RIP is irrelevant to the situation so ignore that. “Internetwork” being the key term in my mind here. So we should have two (2) networks for RouterA-RouterB, then RouterB-RouterC, certainly I wouldn’t waste a /27 on those but in the framework of this question what is one to do?

    After looking closely, 31, 64, 127 and 192 are unusable addresses. This one made me work for the answer.

  23. Q4 Q’s
    January 2nd, 2013

    @anonymous
    Question 4 only has 1 answer, D

    /27 = 255.255.255.224
    Block Size = 32
    Subnet= 192.168.1.32
    Valid Host Range:
    192.168.1.33 – 192.168.1.62

    Thus Host A’s IP and Default Gateway are both in the range, as is SwitchA’s IP.

  24. Fleshwound
    January 9th, 2013

    Man, question 1 answer A is tricky. Took me a few to figure it out. It’s overlapping because different subnets but same address. Tricky

  25. Anonymous
    January 17th, 2013

    Still confused on the first question, Answer A. Answer A states that Fa0/3 has an IP Address that overlaps with 10.1.3.0/30. What it seems like it should say is that Fa 0/3 has IP addresses in different subnets. The wording is incorrect. Not sure how the statement ‘overlaps with 10.1.3.0/30’ makes sense, since there is no /30 and the problem is not overlapping addresses, but instead addresses in different subnets. What am I not understanding here?

  26. Anonymous
    January 17th, 2013

    Still confused on the first question, Answer A. Answer A states that Fa0/3 has an IP Address that overlaps with 10.1.3.0/30. What it seems like it should say is that Fa 0/3 has IP addresses in different subnets. The wording is/seems to be incorrect. Not sure how the statement ‘overlaps with 10.1.3.0/30′ makes sense, since there is no /30 and the problem is not overlapping addresses, but instead addresses in different subnets. What am I not understanding here?

    It seems that for the /25 subnet based on the fa/03 diagram,
    the valid subnets are 10.1.3.0-127 and 10.1.3.128-255. Not sure why considering a /30 for this interface would be relevant based on the diagram.

  27. Retake#ICND1
    January 23rd, 2013

    Hi, I will be retaking the ICND1 exam tomorrow and am pretty ready this time. However, I don’t understand why B is one of the correct answers for Q1. Aren’t the valid subnets for the 10.1.1.0/30 network inclusive of 10.1.1.0, 10.1.1.4, 10.1.1.8…. ? Which would make interface fa0/1 a valid IP address for the subnet in which it resides ????
    Thanks.

  28. Frank
    January 25th, 2013

    @Retake
    B is a right answer because Fa0/3 is on a different network than the one that has been determined. The range of this network is 10.1.3.129-10.1.3.254. Fa0/3 is on the 10.1.3.1-10.1.3.126 network.

  29. Anonymous
    January 29th, 2013

    I signed up for this web site and paid pay pal, but the login does not recognize my user name and password.

  30. Anonymous
    February 12th, 2013

    @Frank
    Actually B is the right answer because subnet 10.1.1.0/30 has an IP range of 10.1.1.0 – 10.1.1.3, therefore the IP address 10.1.1.4 on fa0/1 would be invalid.

  31. Curious
    February 20th, 2013

    I m kinda new at networking….can someone plz tell me what f/a stands for…

  32. Curious
    February 20th, 2013

    Like fa0/3…..what does that means…

  33. Eirik
    February 21st, 2013

    Curious, it stands for “Fast Ethernet”

  34. Anonymous
    February 24th, 2013

    Guys has anyone see the problem that i am seeing on number 1 ANSWER C IS SAYING: C – Interface fa0/2 has an invalid IP address for the subnet on which it resides. This is true becoz the subnet indicates that it has 130 hosts, of the the topology used /25 meaning to say we’re left with 7 bits that will give us 128 hosts when in actual fact we need 130 hosts. So think C could as well be taken as an answer since the explanation on A is wrong or its a typo error. IP Add 10.1.2.126 being used on fa 0/2 is wrong if we’re to accommodate 130 hosts

  35. Anonymous
    February 24th, 2013

    To accomodate 130 hosts we need 8 bits that will give us 256-2=254 so i feel C is one of the Answers. I stand to be corrected

  36. hiba
    March 16th, 2013

    i hope to pass the exam! but how its my dream

  37. Anonymous
    March 21st, 2013

    Although you are correct that /25 cannot accomodate 130 hosts, awser C is still correct since the question states: C – Interface fa0/2 has an invalid IP address for the subnet on which it resides.

    The subnet is /25 that means the broadcast address = .127 since the address on fa 0/2 = .126 its still a valid address within that subnet.

  38. Anonymous
    April 1st, 2013

    any changes on cisco since jan 2013

  39. Paul
    April 11th, 2013

    Any available dumps or guidance please email me at bkboi81@yahoo.com

  40. Achi
    April 17th, 2013

    Hello 9Tut and you all. I am taking my exam next week ICND1 and I would appreciate if you could email me any available dumps.. email: achilleas1979@windowslive.com

    Thank you in advance.

  41. asad usmnai from dubai
    April 24th, 2013

    can u give me answer the what is the valid last address of this ip 192.168.215.227 / 27

    plz help me …i am waiting u.

    thanks

  42. asad usmnai from dubai
    April 24th, 2013

    can u give me answer the what is the valid last host address of this ip 192.168.215.227 / 27
    plz help me …i am waiting u.
    thanks

  43. Broffredo
    April 29th, 2013

    @ asad usmnai The information for the addresses 192.168.215.227/27 is the following:

    Network Address: 192.168.215.224/27
    Valid Hosts: 192.168.215.225 – 192.168.215.254
    Broadcast Address: 192.168.215.255

  44. DM
    May 3rd, 2013

    Q3 was in the exam today

  45. Anonymous
    June 5th, 2013

    @chas yes !!! i just stick to one method of subnetting!

    FIND A METHOD FOR SUBNETTING AND LEARN IT UNTIL YOU CAN ANSWER THE FOLLOWING QUESTIONS:
    -How many borrowed bits? from /8, /16, /24
    -how many Host bits? 32-x (host bits)= y ^2
    -What are my blocks for this CIDR/VLSM- 128 64 32 16 8 4 2 1
    -what is my subnet mask?= .0 .128 .192 .224 .240. 248 .252 .254 .255

  46. sood
    June 25th, 2013

    Can I please get explanation for question 1?

  47. Anonymous
    June 30th, 2013

    Question 1
    Why A? fa0/3 has and ip address of 10.1.3.1 and the network is 10.1.3.0/30 We look at /30 which would give us a subnet of 255.255.255.252 and the magic box number:
    128 192 224 240 248 252 254 255
    128 64 32 16 8 4 2 1
    255 127 63 31 15 7 3 1

    Look at the number below the 252. and you have 4, so the ip block range goes from:
    0|1-2|3
    4|5-6|7
    8 and so on. Now 10.1.3.1 is in the same ip range of 10.1.1.0 so therefore 10.1.3.1 thru 10.1.1.2 is a no go for it overlaps the ip range of 10.1.3.0/30

    Now why B? look at the magic number above is 10.1.1.4 in the same range as 0|1-2|3? nope so it is wrong.

    Why D? This one is easier than you think. Here is how I write it so I am not doing math or counting by 2’s in my head:
    2 /31
    4 /29
    8 /29
    16/ 28
    32 /27
    64 /26
    128 /25
    256 /24
    and so on just do 256*2=X*2 and keep counting down I stop at 8192 /19
    Now we look at the /number and match it to the list above. /25 = 128 – 2 (we are dealing with host so -2) = 126 does this = or >130? No you we need to change it to /24 to give us 256-2=254 host that we can use to meet the 130 host needed.

    I hope this helps

  48. Lateralus
    July 23rd, 2013

    @Anonymous (or anyone else discussing Q1), where are you getting that fa0/3 is a /30? The only interface that’s on a /30 is fa0/1 (10.1.1.0 /30), right?

    The question says fa0/3 is on a /25 (10.1.3.128 /25 to be precise). I completely get that fa0/3’s IP address, 10.1.3.1, IS on a DIFFERENT subnet than it’s hosts are (fa0/3 is on the 10.1.3.0 subnet and the hosts are on the 10.1.3.128 subnet), making it an invalid IP here.

    What I still don’t get is the description of the answer, that “fa0/3 has an IP that OVERLAPs with the network 10.1.3.0 /30” Where is that network? Am I missing something here?

    The kicker to me in all of this is that B, “Interface fa0/1 has an invalid IP address for the subnet on which it resides.” is one of the correct answers here and IS EXACTLY THE SAME SITUATION we’re discussing with A. Option C is a wrong answer, but its worded the exact same way as B, for the same situation also… Seems to me that Cisco got a little too cute here for no real reason. If option A were “Interface fa0/3 has an invalid IP address for the subnet for which it resides.” then there would be little to no confusion.

  49. Raj
    August 7th, 2013

    Question: Show run>>>>>>

    Router(config)#do show ip int br
    Interface IP-Address OK? Method Status Protocol

    FastEthernet0/0 10.1.3.1 YES manual up up

    FastEthernet1/0 10.1.3.1 YES manual administratively down down
    Router(config)#interface fastEthernet 1/0
    Router(config-if)#no shut
    % 10.1.3.0 overlaps with FastEthernet0/0
    FastEthernet1/0: incorrect IP address assignment
    Router(config-if)#

  50. Jayakumar J
    August 16th, 2013

    Is 9tut is under maintenance. ?

    because we can’t open many tutorials. ?
    including subnetting. ?……
    please help me. ?…………

  51. Max
    August 16th, 2013

    I can’t open 9tut.com. 9tut is under maintenance? please help me?

  52. Anonymous
    August 28th, 2013

    latest CCNA 640-802 pass4sure dumps in PDF format is available in http://url.mn/h/5a9ca34

  53. Mncedisi
    September 20th, 2013

    Higuys can someone pls explain how come the answer is: 4092 in Question 2
    and in Question 3 is D Pls I will glad is someone can explain or kindly send an email to this address: mncedisiwiseman5@yahoo.com

  54. Omar CCNP
    September 26th, 2013

    Q3 Explanation….

    All ip addresses given in the option are Network and Broadcast ip addresses. The question asked for an ip address (Usable address), therefore we need to assign it the address that is not a Network nor Broadcast Address.. “0.32.64.96.128.160.192. = .63.95.127.191” <<>>> 190 is the only address that can be use cause is falls in the 160 block as a usable ip.

  55. Will
    September 26th, 2013

    Im taking my ccent today noon..9tut and cbt nuggets. With one on one help from CCNP’s and lots of prayer. Pray i pass!!!

  56. Router47
    October 7th, 2013

    Why is question # answer ‘D’?

    Couldn’t ANY of those subnets have 13 hosts?

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    October 24th, 2013

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  58. Charlotte Hardy
    November 18th, 2013

    We are trying to set up a Class C network with 2 different customers on the same network. I am trying to figure out how to split the 254 hosts between the 2 customers without needing a whole seperate network.

  59. Anonymous
    September 23rd, 2016

    please explain question 3

  60. Router101
    July 21st, 2017

    Question from Anonymous September 23rd, 2016
    We are trying to set up a Class C network with 2 different customers on the same network. I am trying to figure out how to split the 254 hosts between the 2 customers without needing a whole seperate network.
    Answer
    254 is maximum number of ip addresses for a class C address. It cannot be split amongst 2 customers. You can only have one host in this range 1-254

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