Home > CCNA – Subnetting 2

CCNA – Subnetting 2

April 19th, 2015 Go to comments

Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.

Question 1

Explanation

60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires “wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is correct.

Note: Answer D is not correct because 192.168.1.56 is not a network address (/26 -> increment: 64)

Question 2

Explanation

All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is correct.

Question 3

Explanation

We need to summarize 4 subnets so we have to move left 2 bits (22 = 4). In this question we can guess the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different networks. So “/24″ moves left 2 bits -> /22.

Question 4

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= 128 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.

take9bits.jpg

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.

take10bits.jpg

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 5

Explanation

To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (24 = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.

The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0… are different networks).

Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.

Question 6

Explanation

First we should notice that different VLANs must use different sub-networks. In this case Host A (172.16.1.126) and Host B (172.16.1.129) are in different VLANs and must use different sub-networks. Therefore the subnet mask in use here should be 255.255.255.128. In particular, it is 172.16.1.0/25 with 2 sub-networks:

+ Sub-network 1: 172.16.1.0 -> 172.16.1.127 (assigned to VLAN 1)
+ Sub-network 2: 172.16.1.128 -> 172.16.1.255 (assigned to VLAN 2)

-> B is correct.

The IP address 172.16.1.25, which is in the same sub-network with host A so it can be assigned to VLAN 1 -> C is correct.

To make different VLANs communicate with each other we can configure sub-interfaces (with a different IP address on each interface) on the LAN interface of the router -> F is correct.

Question 7

Explanation

“The network administrator needs to address seven LANs” means we have 7 subnets < 8 = 23, so we need to borrow 3 bits from the host part (to create 8 subnets). But the title said “subnet 0 is not being used”, we cannot use the first so in fact we only have 8 – 1 = 7 subnets. We are using class C address block which has 8 bits 0 (the default subnet mask of class C is 255.255.255.0), so the number of bit 0 left is 8 – 3 = 5. Therefore the hosts per subnet will be 25 – 2 = 30 -> E is correct.

Note: There was some confusion here. The title only said “subnet 0 is not being used”, but it did not mention that the command “no ip subnet-zero” is used. Maybe that means we can still use the last subnet (called the All-Ones subnet). In other words, maybe the title implied that “the subnet 0 can be used but the network administrator ignored it for safe”. Thus the last subnet can still be used.

Question 8

Explanation

Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24

Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (22=4).

Question 9

Explanation

We have 4 routes learned by EIGRP:

D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1

These subnets are all /30 (as it says “192.168.25.0/30 is subnetted, 4 subnets”. We have 4 successive subnets = 22 so we can go back 2 bits -> the summarized subnet mask is 30 – 2 = 28 and the summarized network is 192.168.25.16.

Comments (41) Comments
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  1. Anonymous
    August 25th, 2015

    Question 1 : Asks Which SUBNET address should this network use.

    key words here are 60 hosts and SUBNET address
    Both below provide the correct mask for 60 hosts , for the question we want to SUBNET Address.
    .56/26 is the Subnet address .56/26 is a valid host within the .0-.63 block

    192.168.1.64/26 {192.168.1.0- 192.168.1.63}
    192.168.1.56/26 { 192.168.1.64 – 192.168.1.127 }

  2. Addie-
    August 26th, 2015

    Question 7
    Some people say 14 is correct answer. I believe those who say 14 are mistaking the question wording, its not asking for 7 hosts/subnet requiring a /28 16 -2= 14 useable

    its states 7 LANs (7 subnets) = 2^3 ( gives us 8) which means 3 bits is assigned for subnet.
    It’s using class C address, which has by default 24 bits for network.

    8 bits left to be used by the subnet and host (3 already being assigned for subnet),
    That leaves us 5 bits left for host = 2^5 = 32 hosts. S
    ubnet 0 not used first and last available IP are used for network and broadcast,
    This leaves us 30 – 2 = 30 available usable IP addresses.

    Please reply and confimr any thoughts on this?

  3. Addie-
    August 26th, 2015

    Question 7

    The network administrator needs to address seven LANs. RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used. What is the maximum number of usable IP addresses that can be supported on each LAN if the organization is using one class C address block?

    The Actual QUESTION here is—
    “hat is the maximum number of usable IP addresses that can be supported on each LAN ….”

    Maximum number of useable IP addresses..
    that would be maximum # of host /LAN or host per one of the 7 subnets

    its states 7 LANs (7 subnets) = 2^3 ( gives us 8) which means 3 bits is assigned for subnet.
    It’s using class C address, which has by default 24 bits for network.
    8 bits left to be used by the subnet and host (3 already being assigned for subnet),
    That leaves us 5 bits left for host = 2^5 = 32 hosts. S
    subnet 0 not used so first & last IP are used for network and broadcast,

    This leaves us 30 – 2 =
    30 available usable IP addresses. or
    “Maximum # of useable IP addresses”

    in this case answer would be 30

    Please reply and confimr any thoughts on this?

  4. st3fanie
    August 26th, 2015

    Q2-
    I understand the requirement for route summary with NO ADDITIONAL subnets

    so requirement #1 is 10.4 ——

    where i am confused is WHY if its 4 subnets do we move over 2 bits I dont understand how this answer came about.

    Why do we need to move over 2 bits how was this determined?

    I want to understand this not just memorize answers..

  5. st3fanie
    August 27th, 2015

    Question 7: I think the explanation you have here is not correct..

    Question states:
    “…needs to address seven LANs” …… 2^3 = 8 this meets the 7 LAN requirement

    “maximum number of usable IP addresses……..that can support each of the 7 LANS…”
    3 bits borrowed to get 8 Subnets ( LAN’s) Class C block… /27 255.255.255.224
    leaving 5 bits for the hosts === 2^5 32 – 2 = 30 hosts / 8 Subnets(LAN)

  6. 808aloha
    August 27th, 2015

    @ Addie– Q7: Mahalo,for your explanation this makes more sense to me..

    We have to read questions carefully…WATCH the extra wording…with “RIP v1…& subnet 0 not being used.. ” <—-that is all fluff to confuse …

    Cisco will throw a lot of fluff into questions just to confuse us.

    Question 7 is really asking:
    The network administrator needs to address seven LANs. blah,blah,blah….What is the maximum number of usable IP addresses that can be supported on each LAN if the organization is using one class C address block?

    Need 7 LANS
    one Class C Block
    Maximum usable IP address /LAN?

    st3fanie broke it down quite well above..

  7. woazli
    September 5th, 2015

    regarding to question 7:
    Wouldn’t be answer E: 30 be correct?

  8. Anonymous
    September 5th, 2015

    woazil-
    yes 30 would be correct.
    9 tut must have corrected explanation.
    before it stated 14.
    yes 30 is correct.

  9. LV
    September 9th, 2015

    Question 7 still shows that answer ‘C’ is correct instead of ‘E’. (answered 14 but 30 is right)

  10. 808aloha
    September 11th, 2015

    Question 7:
    Answer is 30. see above explnations

    This is an example of why relying on dumps is NOTa substitute for knowing & understanding the concepts & mechanisms behind routing /switching

  11. Asim Roy
    October 17th, 2015

    I see the Question followed by Explanation. But I do not see the statement of the Question. Where may I find all questions meaning the statement of the question.

  12. Asim Roy
    October 17th, 2015

    If someone has all the question, is it possible to email @ asim_roy@yahoo.com. That will be highly appreciated.

  13. omar
    November 1st, 2015

    hi guys, Q2 and Q3, didn’t really explain how the subnet mask was obtained, I finally understood (more or less) how we got that mask, check the following link:

    http://www.orbit-computer-solutions.com/IP-Address—Route-Summarization-Example-_2.php#sthash.3BCrldeF.dpuf

    Although i still dont understand the method used by 9tut here, but using the method in the above link i get the correct answers. Hope that helps.

  14. omar
    November 1st, 2015

    Ok so i figured out the method 9tut used as well:

    Find the default subnet mask being used: for example, classB /24
    Find the total # of subnets being summarized, for example 13 networks, so we’ll use 16 = 2^4
    Summary Mask = 24 – 4 = 20 (or as they say about ‘move left’ 4 bits)

  15. shoogn
    November 23rd, 2015

    can anyone explain Q1 in more details

  16. Yordan
    December 6th, 2015

    Q4
    14 is still the correct answer for me . First and last subnets are not allowed when no ip subnet zero is used or a classful protocol is used like RIP v1. In this case the fact that RIP v1 is used means that that last subnet is not usable too.

  17. Osman
    December 9th, 2015

    please,
    some one tell in Q9 why the summarized network is 192.168.25.16.

  18. foradmin
    January 5th, 2016

    Admin,
    Question 7 correct answer is E – 30
    Explanation is good
    Answer is given wrong as C

    Thank you

  19. q9explained
    January 5th, 2016

    Osman
    Please see explanation for Q9 below
    Q 9

    EIGRP – D Networks Fourth Octet
    192.168.25.16 0001 0000 (16)
    192.168.25.20 0001 0100 (20)
    192.168.25.24 0001 1000 (24)
    192.168.25.28 0001 1100 (28)

    so 0001 XXXX is common for all IPs which is 16 for last octet
    Subnet Mask is 8+8+8+4 = /28 = 240
    So answer is 192.168.25.16/240
    Hope this helps you in understanding Summarization.
    Thank you,

  20. vaibhavgupta
    January 6th, 2016

    I am disappointed because i am not able to see the questions on the screen.

  21. 9tut
    January 6th, 2016

    @vaibhavgupta: You can download the questions at this link: http://www.9tut.com/ccna-questions-and-answers

  22. Waqar @hmad
    January 12th, 2016

    Fastest, easiest subnetting video

    https://www.youtube.com/watch?v=MkTI_O2fRUw

  23. charan
    February 12th, 2016

    Workstation A has been assigned an IP address of 192.0.10.24/28. Workstation B has been assigned an IP address of 192.0.10.100/28. The two workstations are connected with a straight-through cable. Attempts to ping between the hosts are unsuccessful. What two things can be done to allow communications between the hosts? (Choose two)

    A. Replace the straight-through cable with a crossover cable.
    B. Change the subnet mask of the hosts to /25.
    C. Change the subnet mask of the hosts to /26.
    D. Change the address of Workstation A to 192.0.10.15.
    E. Change the address of Workstation B to 192.0.10.111.

    what is the ans and why?

  24. Eli
    March 6th, 2016

    For the professionals.
    Isn’t Q1 misleading? The rules of VLSM says you must first allocate IP and Subnet to the biggest network and then to go down. IE: first 100 users > 50 users > 20 users.
    In this question they start with 30 users > 16 > 16 > and then they request we add 60 users. It’s illogical to the network topology and will cause errors.

  25. aleena
    March 20th, 2016

    i don’t understand question 9 ,can some one explain me ,,thnaku

  26. ManoR
    March 25th, 2016

    @Eli,
    we need to summarize following 4 networks (because these 4 networks were learned by EIGRP)
    Address binary
    ————— -> —————
    192.168.25.16 -> x.x.x.0001 0000
    192.168.25.20 -> x.x.x.0001 0100
    192.168.25.24 -> x.x.x.0001 1000
    192.168.25.28 -> x.x.x.0001 1100
    So the summary address is x.x.x.0001 0000 /28 ( first 4 bits of last octet are common)
    i.e 192.168.25.16/28 ( or mask 255.255.255.240)
    Hope this makes sense

  27. ManoR
    March 25th, 2016

    Sorry, my comment is meant for aleena, not Eli, in relation to Question 9

  28. Anonymous
    April 5th, 2016

    xhcvxcvx

  29. Anonymous
    April 5th, 2016

    good

  30. Anonymous
    April 5th, 2016

    ggddd

  31. Mikky
    April 24th, 2016

    Regarding Question 1
    How come it is not a valid network address? can someone explain?
    Thank you!

  32. Anonymous
    May 9th, 2016

    anyone with CCNA latest dumps help m going to seat for the exam in June

  33. Danny
    May 13th, 2016

    I can’t see questions, Only solutions WHY is that soo?

  34. Dimi
    June 11th, 2016

    Does Question 4 really has a typo??? WTF cisco??? How can we pass the exam??

  35. DoubleA
    July 13th, 2016

    @Dimi

    I second that. “Maximum” is a really bad “typo” (trick) for this question, since it deliberately causes people to think that 255.255.255.224 and 255.255.255.192 are the right answers (even though 192 would still exceed the maximum). Typos like this may be telling us that technicians can’t trust Cisco to clearly communicate exam expectations to them, which is a scary notion. I hate to stir up suspicions, but it means that cisco could be taking people’s money and actually making the test not passable in certain cases – maybe to certain “undesirable” candidates, who may very well be upstanding and capable technicians that nonetheless deserve to pass. This is called creating “a good old boy system” and “rigging the game” and companies can get slammed with huge class action lawsuits for being found as doing that sort of thing to an entire public audience. No evidence on that yet though (I think), so I guess we just keep a lookout for now and hope they’re not that shady.

  36. No.1
    August 7th, 2016

    Can anybody please email me the latest dumps as i am appearing for the exam on 11th august. Thank you.
    (sahil.sehgal60@yahoo dot com)

  37. Anonymous
    October 13th, 2016

    192.168.25.0/20

  38. Lo Sing
    November 15th, 2016

    I thank this helpful community and kind people providing information to others to help pass the exam! I now am CCNA Certified as I passed my CCNA this morning! I used the e8ay link provided by others to pass the exam as well. Gotta say the 139Q dump is accurate! Here is the link if you would like to use: e8ay.com/itm/200-125-v3-Dump-/322325097800? (change the “8” to a “b”)

    Good Luck!

  39. kr
    November 23rd, 2016

    can some plz in understanding Q2

  40. raj
    November 23rd, 2016

    can any one help me with Q2

  41. SJK
    December 4th, 2016

    Where are all the question???I am not being able to find out.. Please help!!

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