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ICND1 – Subnetting

June 15th, 2017 in ICND1 100-105 Go to comments

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

Question 1

Explanation

From the /28 we can find all information we need:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)

In fact we don’t need to find out the broadcast address because the question only asks about subnet address (network address).

Question 2

Explanation

From the /28 we can find all information we need:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 192.168.23.48 (because 48 = 16 * 3 and 48 < 61)

Question 3

Explanation

From the subnet mask of 255.255.255.248 we learn:

Increment: 8 (248 = 11111111.11111111.11111111.11111000)
Network address: 192.168.1.40 (because 40 = 8 * 5 and 40 < 42)

Question 4

Explanation

From the /20 we can find all information we need:

Increment: 16 (/20 = 11111111.11111111.11110000.00000000). This is applied for the 3rd octet.
Network address: 10.1.160.0 (because 160 = 16 * 10 and 160 = 160 -> the IP address above is also the network address.
Broadcast address: 10.1.175.255 (because 175 = 160 + 16 – 1)

Therefore only 10.1.168.0, 10.1.174.255 and 10.1.160.255 are in this range. Please notice 10.1.174.255 is not a broadcast address and can be assigned to host.

Question 5

Explanation

Increment: 32 (224 = 11111111.11111111.11111111.11100000)
Network address: x.x.x.(0;32;64;96;128;160;192;224)
Broadcast address: x.x.x.(31;63;95;127;159;191;223)
-> Last valid host (reduced broadcast addresses by 1): x.x.x.(30;62;94;126;158;190;222) -> Only B is correct.

Question 6

Explanation

Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has been subnetted with a /26 mask so we have 2(26-24) = 22 = 4 sub-networks:

1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)

In all the answers above, only answer C and D are in the same subnet.

Therefore only IPs in this range can be assigned to hosts.

Question 7

Explanation

With network 192.168.20.24/29 we have:

Increment: 8 (/29 = 255.255.255.248 = 11111000 for the last octet)
Network address: 192.168.20.24 (because 24 = 8 * 3)
Broadcast address: 192.168.20.31 (because 31 = 24 + 8 – 1)

Therefore the first usable IP address is 192.168.20.25 (assigned to the router) and the last usable IP address is 192.168.20.30 (assigned to the sales server). The IP address of the router is also the default gateway of the sales server.

Question 8

Explanation

The number of valid host IP addresses depends on the number of bits 0 left in the subnet mask. With a /30 subnet mask, only two bits 0 left (/30 = 11111111.11111111.11111111.11111100) so the number of valid host IP addresses is 22 – 2 = 2. Also please notice that the /30 subnet mask is a popular subnet mask used in the connection between two routers because we only need two IP addresses. The /30 subnet mask help save IP addresses for other connections. An example of the use of /30 subnet mask is shown below:

slash30_subnet_mask.jpg

Question 9

Explanation

Increment: 2 (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0)
Network address: 10.16.2.0 (because 2 = 2 * 1 and 2 < 3)
Broadcast address: 10.16.3.255 (because 2 + 2 – 1 = 3 for the 3rd octet)

-> The lowest (first assignable) host address is 10.16.2.1 and the broadcast address of the subnet is 10.16.3.255 255.255.254.0

Question 10

Explanation

Increment: 4 (/22 = 11111111.11111111.11111100.00000000)
Network address: 172.16.156.0 (156 is multiple of 4 and 156 < 159)

 

Comments (99) Comments
  1. ronnie
    October 31st, 2013

    Thoz dumps are for free the last time I checked so I don’t know why you had to buy

  2. Wanttopass
    November 2nd, 2013

    In Q7 A is also correct – 8 x 1 + 1 etc first host 14 last host 15 broadcast?

  3. french
    November 5th, 2013

    In Q5 there actually 2 good reponses B and C : 61 is in the subnet 192.168.2.32/27 and 62 too

  4. french
    November 5th, 2013

    In Q5 I didn’t see it but D is also in 192.168.2.32/27 : 60 is a valid host address!

  5. french
    November 5th, 2013

    sory I didn’t see the word LAST in the question… the onlyy good answer is B …

  6. Anonymous
    November 17th, 2013

    in question #10 why is 156.0 ….?? I do not understand

  7. Kristof
    November 19th, 2013

    Regarding last comment: the prefix is /22 so the mask in binary is 11111111.11111111.11111100.00000000

  8. Jeol
    December 8th, 2013

    Wanttopass, no, the A is not correct in Q7 because we have to choose the first usable host address for router and the last usable host address for server in subnet 192.168.20.24/29.

  9. Boom
    December 9th, 2013

    @joel, answer is right!

  10. Anonymous
    December 11th, 2013

    In question 7, A and C are correct:

    Answer A
    Network: 192.168.20.8
    Router first usable. 192.168.20.9
    Subnet Mask: 255.255.255.248
    Server last usable: 192.168.20.14
    Broadcast: 192.168.20.15

    Next network: 192.168.20.16

  11. Jeol
    December 11th, 2013

    No. Just look at the Question 7 carefully:
    “For network 192.168.20.24/29, the router is assigned the first usable host address”
    /29 in DDN is 255.255.255.248 = 8 address per subnet.
    192.168.20.24/29:
    Network address: 192.168.20.24
    First usable: 192.168.20.25, this is gateway
    last: 192.168.20.30, this is server
    Broadcast: 192.168.20.31
    Did you got it?

  12. husam
    December 26th, 2013

    please help me i want new question in ccna 120-200
    thanks for all

  13. Anonymous
    January 12th, 2014

    In Q7 explanation….how to know there are 8 subnets from 255.255.255.248 ???

  14. Anonymous
    January 12th, 2014

    sorry, i mean how to know there are 8 addresses per subnet ….thanks

  15. xallax
    January 12th, 2014

    256-248 = 8
    subtract the value of the last incomplete byte from 256 and that’s the short and fast answer

  16. brahmanand
    January 28th, 2014

    hiii,

    In Q 10, how is it 156??

    192 & 128 also are multiple of 4 as well.!!!

    Please explain.

    thanks in advance!!!

  17. CCENT Someday
    January 29th, 2014

    If you need help learning to subnet, etc. Here is a video that I used to refresh and/or learn from:

    http://www.youtube.com/watch?v=rs39FWDhzDs

  18. CCENT Someday
    January 29th, 2014

    Q10: Does not make sense. Please explain in greater detail. Thanks!

  19. CCENT Someday
    January 31st, 2014

    Anyone,

    I plan on taking exam next week. What other sims/labs did you get besides the OSPF 6 router?

    Thanks in Advance!

  20. DaveB
    January 31st, 2014

    The answers for Question 4 are: C. Just that–only C. Read the question carefully:

    Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)

    The key is “for hosts”. They’re asking for host addresses. All other answers are network or broadcast addresses. You can’t choose three, because only one is an address that is valid for a host.

  21. DaveB
    January 31st, 2014

    OK, I take it all back. The light went on: the correct answers are A, C, D. Explanation is good. 2 to the twelfth power (32-20=12), is 16x(2 to the eighth), which means the range is 160.0 through 175.255. Valid host addresses are 10.1.160.1 through 10.1.175.254.

  22. max
    February 19th, 2014

    Q6 ?
    192.168.4.0 / 26
    192.168.4.1
    192.168.4.62
    192.168.4.63

    Why C and D ?

  23. Marco Mwaimu
    February 20th, 2014

    For the one who said A and C are correct for question 7 your very wrong,mind you that you have given the network id of 192.168.20.24/29 so the correct answer was C,the question is so trick so you have to be very carefully ,thanks for the questions they have been important to me

  24. Marco Mwaimu
    February 20th, 2014

    @brahmanand although 192 and 128 are multiple of 4 but are not the networks id for the given host(172.16.159.159/22) so obviously the answer will be 172.16.156.0/22,i hope it has been informative to you friend

  25. Anonymous
    February 22nd, 2014

    question 6

    the subnet ID 0 64 128
    the broadcast network 63 127 255

    so only valid subnets are 65-126
    therefore C. 192.168.4.67
    D. 192.168.4.125

  26. Rasclat
    February 27th, 2014

    are these questions the exact questions on the test? Meaning are the numbers the same here as they’re on the test?

  27. Tried and Failed
    February 28th, 2014

    Even if these are the exact questions on the test, it makes no difference. You need to truly understand subnetting for this test. If they change one octet or even a few numbers, you will have to truly subnet.

  28. Chris
    March 3rd, 2014

    Have a question regarding question 10. I understand that the subnet (Network ID) is 172.16.156.0 and that 172.16.159.159 is part of the subnet with the next subnet being 172.16.160.0 and broadcast 172.16.159.255. So that would give 2 usable addresses 172.16.157.255 and 172.16.158.255. Is my thinking right or am I slightly off course. Any help please as to whether I am on the right path. I do understand that /22 gives 255.255.252.0 and an increment of 4 (-) 2. One address for the Network ID and One for the Broadcast which would give 2 usable addresses. Does the last octet appear as .255 as we have not subnetted into the fourth octet. I think I am on the right path but any help would be appreciated. Thanks Chris

  29. Konan
    March 3rd, 2014

    @Chris
    Simple rule:If the address has 255 at last octet,it doesnt mean that it is broadcast IP.

  30. Anonymous
    March 4th, 2014

    Thanks Konan. So in the above question 10 does that mean we use 255 for the last octet? We subnetted in the 3rd octet. What would be put in the fourth octet? Thanks for any help. Chris

  31. Anonymous
    March 4th, 2014

    I just remember the rule, that if all 0’s then it is the Network ID and if all 1’s then it is the Broadcast.

  32. Don Mega
    March 20th, 2014

    All the answers are as they say they are. Best part about it for me is that I figured them all out in my head. It’s really very simple if you find the right study material. I have only been studying for about a week now. But I do know some things about computers and network. Pertaining to Q7. The subnet is 24 next one starts at 32. 31 is the broadcast so 30 is the last available host in that subnet you can use. Since 24 is the subnet the 1st host is 25. See how easy that was?

  33. ico
    April 1st, 2014

    Hello,

    Would somebody explain question 4.
    Why 10.1.176.1 is couldn’t be a valid host address in 10.1.160/20 network ?In my opinion 10.1.176.1 is a part from 10.1.176.0 – 10.1.191.255 network {10.1.176.1 – 10.1.191.254 valid host}

  34. Ron
    April 3rd, 2014

    @ico,
    Q4
    Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)

    /20 = 16 block size
    160 = multiple of 16
    so you would think the host range would be 10.1.160.1 to 10.1.175.255 BUT you need to take away the broadcast address.

    so the real range of hosts is 10.1.160.1 to 10.1.175.254..
    and the next network (block size 16!) is 10.1.176.0

    thus the answers being A, C, D

    A. 10.1.168.0 (valid host)
    B. 10.1.176.1 (in the next subnet!)
    C. 10.1.174.255 (valid host)
    D. 10.1.160.255 (valid host)
    E. 10.1.160.0 (network address)
    F. 10.1.175.255 (broadcast address)

  35. ico
    April 4th, 2014

    Thank you Ron. I got it.
    There was my mistake.

  36. Dx
    April 6th, 2014

    I just want to ask if in the exam they are going to have questions that are large numbers to subnet like 186.218.196.42/20? The increment is 16 here

  37. shyamal
    April 12th, 2014

    Answer to question 7 is C. Read the question properly, it is asking For network 192.168.20.24/29….No the subnets that are created using 0,8,16,24,32…..ETC.
    For network 192.168.20.24 Subnet the first usable is .25 and last is .30 (.24 is network and .31 is broadcast).
    Hope this help to understand.

    Harsh

  38. mido
    April 23rd, 2014

    I do not understand question7

  39. fez
    April 24th, 2014

    Passed my 100-101 exam today and question 1,2,3,4,5 and 6 were there.. all the best

  40. GregPL
    April 27th, 2014

    on Q6 , the 4th subnet range should be
    4th subnet: 192.168.4.192 (to 192.168.4.255)????
    the explanation might confuse people please fix this! it is a /26 so the increment is 64
    Currently you have posted
    1st subnet: 192.168.4.0 (to 192.168.4.63)
    2nd subnet: 192.168.4.64 (to 192.168.4.127)
    3rd subnet: 192.168.4.128 (to 192.168.4.191)
    4th subnet: 192.168.4.192 (to 192.168.4.225)

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    May 16th, 2014

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  42. Saavy Boy
    May 17th, 2014

    Question Six .125 does not fall under .127 there is a mistake there.

  43. Saavy Boy
    May 17th, 2014

    Nevermind it does fall under. so it’s OK

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    May 29th, 2014

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  46. Jack
    June 3rd, 2014

    What is the network address for the host with IP address 192.168.23.61/28?

    A. 192.168.23.0
    B. 192.168.23.32
    C. 192.168.23.48
    D. 192.168.23.56
    E. 192.168.23.60

    I think answer B. If increment is 16….32…64..128…. Why 48 ?…………….. Please explain

  47. hary
    June 3rd, 2014

    can any 1 explain question 10…..

  48. Neil
    June 3rd, 2014

    @Jack
    Q2
    Yes increment is 16. So network will be divided as below.
    192.168.23.0 – 192.168.23.17
    192.168.23.16 – 192.168.23.31
    192.168.23.32 – 192.168.23.47
    192.168.23.48 – 192.168.23.63 <——–
    192.168.23.64 – 192.168.23.79

    Host ip .61 belongs to network ranging from .48 – .63
    .48 being the network address

    Hope it helps.

  49. Neil
    June 3rd, 2014

    ^typo error 192.168.23.17 should be 192.168.23.15

  50. tut
    June 8th, 2014

    guys check this link… Very good video on ip addressing and sub netting

    https://www.youtube.com/watch?v=FrdaCx3R9CM

  51. Anonymous
    July 9th, 2014

    Please may you explain in detail question no 9?

  52. Tark
    July 31st, 2014
  53. Tark
    July 31st, 2014

    /x = CIDR Mask
    Blue # = DDN mask in interesting octet.
    Green # = Magic Number

    Remember this chart for exam if you have trouble working the mental math out.

  54. anon
    August 6th, 2014

    Anyone seen any VLSM questions on the exam?

  55. baa
    September 1st, 2014

    q4? 10.1.160.0 is not valid host. its a network address

  56. Dima
    September 25th, 2014

    Question 2: What is the network address of 192.168.23.61/28?
    A bit misleading… The network address is 192.168.23.0
    As for the subnet address, then 192.168.23.48

  57. Soiree
    October 7th, 2014

    Hi Fox,
    I need a help on the way i could respond on this type of subnetting question in less than 2 minutes: What is the subnet address of 172.16.159.159/22?
    I know that this is a class B address and the default mask is 255.255.0.0
    Also, I know that /22 is 255.255.252.0 subnet mask and the magic number is 4
    what can be the fastest way to found out that 159 on the third and fourth octect is a broadcast address without going by counting the increment 4(0,4,8,12,16….156,160…). Thanks for your help.

  58. Lee
    October 8th, 2014

    @Soiree found this on http://www.subnetting.net/TrickShots.aspx

    Q: Suppose I have a block size of 8 and I’m trying to find the IP range for the number 122. Is there a faster way to get there than by starting at 0, and counting by 8 until I find 120 and 128 to know the range? During the exam this will take too long! -Asked by Esrar from San Jose, CA

    A: Great question! Yes, there is a faster way. Once you know the block size, you can divide the IP number by the block size to find the multiplier: 122/8 = 15.something. In this case the division results in 15.25 but the number to the right of the decimal doesn’t matter, which is why we wrote it as “15.something”. Now take the integer portion of that answer (15) and multiply it by 8, which gives you 120, which is the starting point you are looking for.

  59. Soiree
    October 9th, 2014

    Thanks you very much Lee for the clue in subnetting questions.
    Good and fast way to subnet.
    Best regards,

  60. Loh
    November 6th, 2014

    i dont understand question number 8

  61. Rodrigo Planche
    November 24th, 2014

    The Answer of Q10 is Wrong.

    Question 10
    What is the subnet address of 172.16.159.159/22?

    A. 172.16.0.0
    B. 172.16.128.0
    C. 172.16.156.0
    D. 172.16.159.0
    E. 172.16.159.128
    F. 172.16.192.0

    Answer: C – WRONG!!!!

    Explanation:

    IP: 172.16.159.159/22 or 255.255.248.0 – 256-248 = 8 in 8 on the third octet.

    156 is not a multiple of 8.

    Subnet:
    172.16.152.0 – 172.16.159.255

    IP 172.16.159.159 belongs to 172.16.152.0/22 subnet. You can check on online ip subnet calculator.

    Becareful!

  62. Rodrigo Planche
    November 24th, 2014

    Sorry my comment above. Is /22 not /21 Q10 – Answer C is right.
    Sorry again.

  63. Umer
    December 1st, 2014

    Q 10 has 172.16.159.159/22
    /22 means 255.255.252.0
    as there are six bits on in second octet so 2exp6=64 subnets and 11111100=252

    now total of 64 subnets ….256-252=4 so 0,4,8,12 so on so multiple of 4
    and 159.159 subnet should be 172.16.156.0

    hope it helps

  64. furmal
    December 9th, 2014

    my test is on 22nd of dec, any body wants to group study for subnetting ?

  65. Deriq
    December 15th, 2014

    Sure. How do we do it. I am doing the exam on Thursday tho

  66. ananias
    December 20th, 2014

    @deriq
    mine is on monday 29th, ccent brainbump from anyone? appreciations.

  67. ananias
    December 20th, 2014

    pls email braindump to abrahamshen@rocketmail.com Thanks-a-million!!

  68. ciscoman
    January 6th, 2015

    Q10 is still confusing .. im using the method jeremy taught on cbtnuggets anyone doing it that way? Please help. Thanks.

  69. ananias
    January 6th, 2015

    test

  70. Islam Is the Best
    January 9th, 2015

    MONDAY HERE I COME ICND1

  71. tara
    January 10th, 2015

    hi
    can i have a CCENT material and which one is eaiser CCENT or CCNA
    thanks

  72. WTF
    January 12th, 2015

    Tara are you trolling or just retarded??

  73. cktx
    January 14th, 2015

    LOL Tara and WTF

  74. Slappypappyj
    January 17th, 2015

    Q4 isnt the first valid ip for .160 network .160.001? So 160.255 a vaild ip? Im might be missing something.

  75. tara
    January 19th, 2015

    any ccent practice test questions

  76. Techgique
    January 26th, 2015

    For those asking about Q10:

    What is the subnet address of 172.16.159.159/22?

    A. 172.16.0.0
    B. 172.16.128.0
    C. 172.16.156.0
    D. 172.16.159.0
    E. 172.16.159.128
    F. 172.16.192.0

    When looking at the octets, know where the /masks are
    (/8) (/16) (/24) (/32)
    (172) . (16) . (159) . (159)

    So when we see a /24, we are dealing with the fourth (last) octet. HOWEVER, in this question it asks for /22 so we are in between /16 and /24, meaning we are dealing wit the third octet. Most of you know your fourth octet subnet sizes but get confused when it goes to octets prior to that. Think of a /22 as a /30, which we know has a subnet size of 4.

    So if we know that the subnet size is four, that’s a little easier to deal with. Now we can (for now) ignore the fourth octet and just look at the increments of four to see our network addresses:

    172.16.0.x
    172.16.4.x
    172.16.8.x
    …all the way down to
    172.16.156.x
    172.16.160.x

    each of these subnets have four numbers in the third octet but they also have 255 numbers in the fourth octet as they count up

    Subnet #1:
    172.16.0.1
    172.16.0.2 etc

    Subnet #2:
    172.16.4.1
    172.16.4.2 etc

    We know that our address given is 172.16.159.159, so if we ignore (temporarily) the fourth octet, we can see that 172.16.159.x falls in between 172.16.156.x and 172.16.160.x

    172.16.156.0 <— (The subnet address)
    172.16.156.1
    172.16.156.2
    …all the way down to
    172.16.159.158
    172.16.159.159 <— (Here it is in the same subnet!)
    172.16.159.160
    …all the way down to
    172.16.159.255 <—(The broadcast address of the subnet)

    Hope this makes a little more sense, as dealing with octets other than /24 and above can get confusing.

  77. Nobody
    February 27th, 2015

    Q4…baa and slappypappyj…..remember every subnet in this example has 4094 hosts….that’s the trick and what is throwing both of you…it’s easier if you write this one out in binary

  78. Ahmed
    March 28th, 2015

    q4,8 in icnd1 exam

  79. conven
    April 9th, 2015

    Can someone please tell me how many TOTAL subnetting questions we can expect in exam? 5? 10?

  80. Nino
    May 14th, 2015

    hi!
    Could you please clarify on Q3 – why the answer is C?

  81. Anonymous
    May 20th, 2015

    @Nino
    Q3 Because 255.255.255.248 248 Mask is a /29

    /29 = 8 hosts ( 6 valid) so the IP address is a .42 so we know its part of the .40 subnet

    answer is C:. 192.168.1.40/29

  82. ChadRex
    May 20th, 2015

    @Nino
    Q3 Because 255.255.255.248 = 248 Mask is a /29
    /29 = 8 hosts ( 6 valid) so the IP address is a .42 so we know its part of the .40 subnet
    answer is C:. 192.168.1.40/29

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    June 24th, 2015

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  86. Goose
    July 10th, 2015

    Q6 (Which two addresses can be assigned to hosts within the SAME subnet?), aren’t C & D in two separate subnets, I believe the question refers only to subnet 192.168.4.0/26 which involves a range of addresses of 4.1 to 4.62, only A and B fall within this subnet but B is a broadcast so not usable for hots, the rest are all in different subnets, unless the question refers to 192.168.4.0/24 then all of them are part of that network.

  87. 9tut
    July 13th, 2015

    @all: We had to move all the questions and answers out of 9tut. We can only keep the explanation. You can download the questions and answers at: https://mega.co.nz/#!oIdESYbD!yyu33vygrfKPy4rcmcbV6qW2fxINNoTokuDM3CjA_og

  88. Liliya
    July 16th, 2015

    Hi everyone! How come i do not see questions for ICND2.. My membership might have expired but i can’t get to my profile to re new it..

  89. 9tut
    July 16th, 2015

    @Liliya: You can access your profile at http://www.9tut.net/member/profile

  90. Hadrami
    July 27th, 2015

    hi all.
    I took ICND1 23-07-2015 and passed 907/1000………….thanks 9tut. I only study this site and got 907/1000 so, if you study extra sites you will be better than me. yeh there is no any configuration only practice “show COMMANDS” and practice all the SIMs.

  91. JB Blanca
    August 13th, 2015

    Hi,
    Just passed the exam last week and there was a lot of subnetting, mostly Class A and B. Please watch this video because it did help me;
    https://www.youtube.com/watch?v=rFra2TtBrk4

  92. preety
    November 30th, 2015

    how can i get question of dis ans???? plzz help

  93. charles
    January 8th, 2016

    hi guys can help me with this ,Network address: 192.168.1.40 (because 40 = 8 * 5 and 40 < 42) why and how to get the '8 * 5' ????? why *5 ?? Q3

  94. charles
    January 8th, 2016

    Question 6

    Explanation !!! can’t understand at all pls help out guys

  95. subnet
    January 21st, 2016

    guys for question 4 I am getting broadcast as 10.0.191.0/20
    Can someone please help where I am going wrong? I got 10111111 as the 3rd octet

  96. Bossaver
    April 25th, 2016

    @charles. I too was confused until I took a triple look. There are four possible subnets that can be created.

    192.168.4.0/26
    192.168.4.64/26
    192.168.4.128/26
    192.168.4.192/26

    What the question is basically asking is, which of these four subnetworks holds two of these hosts addresses. The answer is C & D because the host range of the 192.168.4.64/26 network is 192.168.4.65-.126/26

    Hope this helps you or anyone else that is confused.

  97. dd
    May 31st, 2016

    please can someone send me the latest icnd1 and icnd2 dumps to davakaniruatgmaildotcom

  98. erict21
    June 25th, 2016

    I really am confused here:
    “An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two)”

    It states 192.168.4.0 ! (so not: 192.168.1.64). So in my honest opinion, the question is noted wrong…
    Am I correct?

  99. erict21
    June 25th, 2016

    ow, never mind, already see the comment of Bossaver now 🙂