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Subnetting Questions

November 6th, 2018 Go to comments

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

Question 1

Explanation

Network A needs 120 hosts < 128 = 27 -> Need a subnet mask of 7 bit 0s -> “/25″.

Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.

Answer E “Link A – 172.16.3.40/30″ is not correct because this subnet belongs to MARKETING subnet (172.16.3.32/27).
Answer F “Link A – 172.16.3.112/30″ is not correct because this subnet belongs to ADMIN subnet (172.16.3.96/27).

Question 2

Explanation

Although all above answers are correct but 172.16.1.0/26 is the best choice as it is the most specific prefix-match one.

Question 3

Question 4

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 5

Explanation

From the subnet mask of 255.255.255.240 (/28) we learn there are 24 – 2 = 14 hosts per subnet.

Question 6

Explanation

From the subnet mask of 255.255.248.0 we learn that the increment is 8 therefore 172.16.8.0 is a network address which cannot be assigned to a host. Other network addresses are 172.16.16.0, 172.16.24.0, 172.16.32.0… Notice that 172.16.31.0 is a valid host address (which belongs to 172.16.24.0 to 172.16.31.255 subnet).

Question 7

Explanation

“/25” means 1111 1111.1111 1111.1000 0000 in binary or 255.255.255.128 in decimal.

Question 8

Question 9

Explanation

The principle here is if the subnet mask makes two IP addresses 10.1.0.36 and 10.1.1.70 in the same subnet then the Network device A does not need to have IP addresses on its interfaces (and we don’t need a Layer 3 device here).

A quick way to find out the correct answers is notice that all 255.255.255.x subnet masks will separate these two IP addresses into two separate subnets so we need a Layer 3 device here and each interface must require an IP address on a unique IP subnet -> A, C are not correct while B, D are correct.

With 255.255.254.0 subnet mask, the increment here is 2 in the third octet -> the first subnet is from 10.1.0.0 to 10.1.1.255, in which two above IP addresses belong to -> each interface of Network device A does not require an IP address -> E is correct.

Question 10

Question 11

Explanation

The RFC 1518 is Classless Interdomain Routing (CIDR). CIDR is a mechanism developed to help alleviate the problem of exhaustion of IP addresses and growth of routing tables.

The problems were:

+ With the classful routing system, individual networks were either limited to 254 hosts (/24) or 65,534 hosts (/16). For many network enterprises, 254 hosts were not enough and 65,534 were too large to be used efficiently.
+ Routing information overload. The size and rate of growth of the routing tables in Internet routers is beyond the ability of current software (and people) to effectively manage.
+ Eventual exhaustion of IP network numbers.

To solve these problem, CIDR was selected as the solution in 1992.

In contrast to classful routing, which categorizes addresses into one of three blocks, CIDR allows for blocks of IP addresses to be allocated to Internet service providers. The blocks are then split up and assigned to the provider’s customers.

According to the CIDR standard, the first part of an IP address is a prefix, which identifies the network. The prefix is followed by the host identifier so that information packets can be sent to particular computers within the network. A CIDR address includes the standard 32-bit IP address and also the network prefix. For example, a CIDR address of 200.1.45.2/26, the “/26” indicates the first 26 bits are used to identify the unique network, leaving the remaining bits to identify the specific hosts.

Therefore, instead of assigning the whole block of a class B or C address, now smaller blocks of a class can be assigned. For example, instead of assigning a whole block of 200.1.45.0/24, a smaller block, like 200.1.45.0/27 or 200.1.45.32/27, can be assigned.

Comments (36) Comments
  1. jasin
    March 6th, 2017

    You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. Which network address and subnet mask meet this requirement?
    A. 192.168.252.0 255.255.255.252
    B. 192.168.252.8 255.255.255.248
    C. 192.168.252.8 255.255.255.252
    D. 192.168.252.16 255.255.255.240
    E. 192.168.252.16 255.255.255.252
    any one please could explain how we goina work out for 3 web server

  2. skyflakes
    March 7th, 2017

    We want to have 3 host right or 3 web servers? so B is correct because of the subnet. 255.255.255.248 or /29. the increment of it is 8. 8 – 2 = 6 usable host, so it would fit the requirements,

  3. Aaron
    March 7th, 2017

    8 – 2 because we subtract the network id and broadcast id. So we could only use 6 host, unlike in letter C for instance is 255.255.255.252. so it’s 30. the increment is 4 so 4 – 2 = 2 host only, so it wouldn’t match the needed requirements for 3. i hope my answer is correct. :)

  4. Eriol
    March 9th, 2017

    Both Skyflakes and Aaron are correct (: because you need to use a mask of /29 to have 6 usable IP’s to configure your “3” needed host, the masks /30 only gives you 2 usable IPs which doesn’t meet the requirements of 3 hosts and the /28 mask gives you 14 usable IPs, which doesn’t fullfil the “while providing the maximum number of subnets” statement.

  5. Hicham
    April 4th, 2017

    Sorry i don’t see quetions???

  6. Henchman
    April 7th, 2017

    Just at first glance of your answers I would say either ” B ” or ” C ” could be used, but then I had to look at your Subnet Mask in the last octet, and knowing that I need to provide for 3 Web Servers and give them IP Addresses from my subnet mask, I would have to borrow 5 bits from left to right giving me a mask ending with 248, i.e 255.255.255.248, and the 5 borrowed bits will give me 8 host minus the broadcast and IP so I will have 6 usable IP addresses for the 3 Web Servers leaving me with 2 left over IP for other devices……that is how I would do it !!!!!

  7. Vince
    April 30th, 2017

    Why do you say subnet hosts. This kind of vocabulary is common among people who don’t know how the technology is used correctly.

  8. Ricky
    May 13th, 2017

    Need help#
    Suppose You have a big company of 63 hosts per network. You are the System Administrator of your company. You have to plan the addressing scheme. Your are given IP 172.72.157.253 And your Dept. (Engtneering) uses Subnet N 399.
    Q. What will be the Subnet bit?

  9. MUhammad Mohyuddin
    May 16th, 2017

    You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. Which network address and subnet mask meet this requirement?

    B. 192.168.252.8 255.255.255.248

  10. davegi
    May 22nd, 2017

    why the payment method is only paypal? any other options like mastercard and others

  11. Thaddeus
    June 28th, 2017

    Confirming the 747q dumps as valid, you can find them here https://drive.google.com/open?id=0B5mAFqgydmCza1ViTmFlNkQwYzg

  12. Anonymous
    June 30th, 2017

    the question doesn’t appear,what’s the problem?

  13. idiott
    July 16th, 2017

    @Matt

    Its appeared only 10 question what about remaining ?
    It shows only demo version… pls send me full version aktparthiban @ gmail com

  14. Anonymous
    August 21st, 2017

    The dumps from IT-Libraries are still valid there are a lot of people recommending them and it was worth it.
    Found the link here http://congressreiki.ranm.org/?all=access-list-questions

    Thanks tut, on to ccnp now

  15. anon
    September 10th, 2017

    Question 10

    A & B are equally valid answers.

    Either change the subnet mask to 255.255.255.224 or smaller
    or change the gateway address so the last octet is between 25-33.

    either will fix the problem.

  16. Question 10 -Anon
    October 24th, 2017

    The reason why the answer to Question 10 is A is because the subnet of the host is .240, on a 192.168.10.16 network. The usable IP’s are 17-30 (31 is broadcast, and 32 is the next network). The gateway is on 192.168.10.1. You need L3 to get to it. Answer B “The default gateway address for Computer A is incorrect” refers to the “gateway” address on the host…no the IP of the router.

  17. Adam
    November 8th, 2017

    Subnettning is very important… we need more examples;)
    you can find more examples also here
    http://ipcisco.com/ip-subnetting-subnetting-examples/

  18. nakajims
    December 14th, 2017

    Question 11 is asking RFC1918 which is IPv4 private address. But explanation mentiond RFC1518.??

  19. mohamoud sh
    January 8th, 2018

    please can you help me today i want to attend an exam about network adminstrator

  20. cthelite
    January 17th, 2018

    RFC 1518 is considered to be the first solution to IPv4 address exhaustion… it created CIDR… but didn’t solve any issues with the growing global routing tables, so answer “A” only would apply to RFC 1518…

    RFC 1519 specifically dealt with CIDR AND Address Assignment and Aggregation Strategies, so if that was the question, “A” and “D” would likely be the answer…

    But the question is about RFC 1918 and neither of these previous RFCs did anything to increase the number of publicly attachable hosts beyond what can be accommodated by a 32-bit addressing scheme, which was the ultimate limit to IPv4.

    So RFC 1631 (NAT) and subsequently 1918 (Private Addresses) did what the other two could not… now you could scale the number of nodes attached to the Internet to literally trillions of devices… still a workaround, a bandaid, but this arguably did the most to alleviate the impending IPv4 address exhaustion…

    Sooo, “A” and “B” dealing with IP addresses and their limitation (running out and overlapping) are the two most correct answers for the two reasons for RFC 1918.

    “C” is obviously not correct as it pertains to IPv6… and as discussed “D” would apply more to RFC 1519… “E” is not a good answer because private addresses don’t “support” NAT, but rather NAT supports the use of private addresses on the internet.

    So best answer is “A” and “B”.

  21. Q9
    April 2nd, 2018

    Guys, on Q9, the answer E is absolutely incorrect. It´s impossible to configure a masc without an IP address. There is only 2 correct answers (B&D), when the question asks for 3.
    Please, review this question. Thanks

  22. Q9
    April 2nd, 2018

    Regarding Q9, the exhibit shows L2 switches, in which, it is impossible to configure ip addresses and masks on their interfaces, therefore, it would never route the networks.
    The answers don´t seem consistent with the exhibit has A,B,C & E are not representing what the exhibit is showing (only D is). Either way, there needs to be a router mechanism to make them communicating. I´m pretty sure Q9 is asking for only 1 answer, not 3. (This is the only way that the question makes sence.)

  23. Satchmo
    April 9th, 2018

    Do you have to be a premium member to view the questions

  24. Muhiz
    April 14th, 2018

    @satchmo did u get the questions ???

  25. 9TUT Request Help with Question #6
    April 28th, 2018

    Question 6
    Assuming a subnet mask of 255.255.248.0, three of the following addresses are valid “HOST ADDRESSES”. Which are these addresses? (Choose three)

    My reference illustrating (network-broadcast)
    (0-7) (8-15) (16-23) (24–31) (32-39)

    A. 172.16.9.0 {Agree/Yes}
    B. 172.16.8.0 {Network Address Answer = No}
    C. 172.16.31.0 {Broadcast Address Answer = No}
    D. 172.16.20.0 {Agree/Yes}

    9TUT Answer: A C D

  26. 9TUT Request Help with Question #6
    April 28th, 2018

    Sorry, I figured out my question above…
    172.16.24.1 – 172.16.31.254

  27. AB
    May 2nd, 2018

    @9tut – Can you please add this explanation as alot of us are confused with the answer chosen.

    Question 3
    You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. Which network address and subnet mask meet this requirement?

    B. 192.168.252.8 255.255.255.248

    Answer:
    The question asks for 3 web severs which means only /29 (.248) & /27 (.240) masks are the only valid subnet masks. since other masks doesnt have 3 usable IP’s for these web-servers.

    The reason why .248 mask was chosen instead of .240 is because the question asks for providing MAXIMUM number of subnets. The network bit for .248 (/29) ends on the 5th digit on the subnet mask (on 4th octet) which brings the number of subnet to 2 to the power of N (N=network bit) which is 2^5=32.
    P.S: For .240 mask, number of subnet would be 2^4 (since the network bit ends on the 4th digit) which = 16. Hence the answer is B with .248 mask.

  28. Anonymous
    May 31st, 2018

    For qn 9E, the subnet is 10.1.0.0/23
    and usable host range: 10.1.0.1 – 10.1.1.255
    Hence, both 10.1.0.36 and 10.1.1.70 are in the same subnet. So, each interface does not require an IP address.

    Answer E is correct

  29. KroLogical
    June 25th, 2018

    Guys, about QUESTION 426 again:
    For which two reasons was RFC 1918 address space define (Choose two)

    D. reduce the size of ISP routing tables

    Waht about this answer, when using private IP addresses (which are not routable on the internet), the ISPs indirectly reduce the size of their routing tables, right ?

  30. HElp
    August 17th, 2018

    i cant see the question

  31. where are the questions guys.. only explanation is there.
    September 6th, 2018

    where are the questions guys.. only explanation is there.

  32. Sonpham
    September 21st, 2018

    @HELp, You need to find on another page in this website

  33. Zed
    October 14th, 2018

    In Q2, how to know the correct routing match to reach a network in another match? is /26 was chosen because it’s the closest subnet to /32 than /24 & /25 ???

    please someone answer me

  34. Zed
    October 14th, 2018

    sorry I meant (((to reach a network in another mask**))))

  35. Hashi
    October 24th, 2018

    What is the correct routing match to reach 172.16.1.5/32?

    A. 172.16.1.0/26
    B. 172.16.1.0/25
    C. 172.16.1.0/24
    D. the default route

    Answer: A

    I agree you Zed there must b wrong something with this question is not clear prefix /32 then they said the correct answer is /26

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