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Subnetting Questions

May 11th, 2017 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

Comments (16) 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

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    May 22nd, 2017

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

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    June 28th, 2017

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  12. Anonymous
    June 30th, 2017

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    July 16th, 2017

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    August 21st, 2017

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  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.

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