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CCNA – Subnetting

January 10th, 2011 Go to comments

Here you will find answers to CCNA Subnetting Questions

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

Question 1

Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)

A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87
E – 201.45.116.159
F – 217.63.12.192


Answer: B C D

Explanation

A subnet mask of 255.255.255.224 has an increment of 32 (the binary form of the last octet is 1110 0000) so we can’t use numbers which are the multiples of 32 because they are sub-network addresses. Besides, we can’t use broadcast addresses of these sub-networks (the broadcast address of the previous subnet is calculated by subtracting 1 from the network address). For example the network address of the 2nd subnet is x.x.x.32 then the broadcast address of the 1st subnet is 32 – 1 = 31 (means x.x.x.31).

By this method we can calculate the unusable addresses, which are (notice that these are the 4th octets of the IP addresses only):

+ Network addresses: 0, 32, 64, 96, 128, 160, 192, 224.
+ Broadcast addresses: 31, 63, 95, 127,159, 191, 223.

Question 2

Which of the following host addresses are members of networks that can be routed across the public Internet? (Choose three)

A – 10.172.13.65
B – 172.16.223.125
C – 172.64.12.29
D – 192.168.23.252
E – 198.234.12.95
F – 212.193.48.254


Answer: C E F

Explanation

Addresses that can be routed accross the public Internet are called public IP addresses. These addresses belong to class A, B or C only and are not private addresses.

Note:

Private class A IP addresses: 10.0.0.0 to 10.255.255.255
Private class B IP addresses: 172.16.0.0 to 172.31.255.255
Private class C IP addresses: 192.168.0.0 to 192.168.255.255

Class D addresses are reserved for IP multicast addresses and can’t be routed across the Internet (their addresses begin with 224.0.0.0 address).

Also we can’t use 127.x.x.x address because the number 127 is reserved for loopback and is used for internal testing on the local machine.

Question 3

A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two)

A – 255.255.255.0
B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
E – 255.255.255.192
F – 255.255.248.0


Answer: B E

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 27-2 = 126 > 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 4

Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)

A – 115.64.8.32
B – 115.64.7.64
C – 115.64.6.255
D – 115.64.3.255
E – 115.64.5.128
F – 115.64.12.128


Answer: B C E

Explanation

CIDR stands for Classless In4ter-Domain Routing, the difference between CIDR and VLSM is slim and those terms are interchangeable at CCNA level.

To specify which IP addresses fall into the CIDR block of 115.64.4.0/22 we need to write this IP address and its subnet mask in binary form, but we only care 3rd octet of this address because its subnet mask is /22.

CIDR.jpg

(x means “don’t care”)

Next, we have to write the 3rd octets of the above answers in binary form to specify which numbers have the same “prefixes” with 4.

4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100

We can see only 7, 6 and 5 have the same “prefixes” with 4 so B C E are the correct answers.

Question 5

Refer to the diagram. All hosts have connectivity with one another. Which statements describe the addressing scheme that is in use in the network? (Choose three)

VLAN_IP_Assignment.jpg

A – The subnet mask in use is 255.255.255.192.
B – The subnet mask in use is 255.255.255.128.
C – The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D – The IP address 172.16.1.205 can be assigned to hosts in VLAN1.
E – The LAN interface of the router is configured with one IP address.
F – The LAN interface of the router is configured with multiple IP addresses.


Answer: B C F

Explanation

VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.

By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) and can be assigned to hosts in VLAN 1.

For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into multiple sub-interfaces with multiple IP addresses -> F is correct.

Question 6

The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet if the ip subnet-zero command is configured on the router? (Choose three)

A – 172.25.78.243
B – 172.25.98.16
C – 172.25.72.0
D – 172.25.94.255
E – 172.25.96.17
F. 172.25.100.16


Answer: A C D

Explanation

If the “ip subnet-zero” command is configured then the first subnet is 172.25.0.0. Otherwise the first subnet will be 172.25.32.0 (we will learn how to get 32 below).

The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000).

First subnet: 172.25.0.0/19
Second subnet: 172.25.32.0/19
Third subnet: 172.25.64.0/19
4th subnet: 172.25.96.0/19
5th subnet: 172.25.128.0/19
6th subnet: 172.25.160.0/19
7th subnet: 172.25.192.0/19
8th subnet: 172.25.224.0/19

In fact, we only need to specify the third subnet as the question requested. The third subnet ranges from 172.25.64.0/19 to 172.25.95.255/19 so A C D are the correct answers.

Question 7

Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from router A?

VLSM_summary.jpg

A. 172.16.0.0/16
B. 172.16.0.0/20
C. 172.16.0.0/24
D. 172.32.0.0/16
E. 172.32.0.0/17
F. 172.64.0.0/16


Answer: A

Explanation

Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.

All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.

The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.

-> Only answer A has these 2 conditions -> A is correct.

Comments
Comment pages
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  1. Lakmal Achira Perera
    January 28th, 2013

    I think it is easy to work with prefix length……try to master below method, I would say in my experience that you can easily find the answer for sub-netting questions

    50 host
    128 64 32 16 8 4 2 1
    1 1 1 1 1 1
    6 bits for host side so therefore 32-6 = 26 (255.255.255.192)

    300 subnet
    512 256 128 64 32 16 8 4 2 1
    1 1 1 1 1 1 1 1 1
    9 bits for network side therefore 16+9 = 25 (255.255.255.128)
    note- in addition, it says working with one class b address therefore class B default mask is 255.255.0.0 (prefix length for class b address is /16)

  2. Lakmal Achira Perera
    January 28th, 2013

    50 host
    increment should be 32 so we put a flags on 32,16,8,4,2,1 -6 bits

    300 subnet
    increment should be 256 and we put a flags on 256,128,64,32,16,8,4,2,1 -9 bits

    sorry for the typing mistake which i was comment above…..

  3. RYAN
    January 30th, 2013

    B AND E IS THE RIGHT ANS. FOR QUESTION #3

  4. route summarization
    January 30th, 2013

    @9tut: do we have route summarization topic here?

  5. 9tut
    January 30th, 2013

    @route summarization: Route summarization is the reverse with subnetting so you can deduce it from subnetting process.

  6. route summarization
    January 31st, 2013

    @9tut: i’m bit confused. can you give me an example please? thanks…

  7. Rahul
    February 1st, 2013

    Thank u 9tut……cleared my exam and scored 920….all questions from 9tut and Shekhars dump…sims was ACL2,EIGRP and Implmntn sim.

  8. Akshat
    February 6th, 2013

    Guyz according to me Q3. answer should be only B. 255.255.255.128
    Am i right or not. Pls Tell me

  9. BigGuy5002
    February 8th, 2013

    @Akshat… the question is not clear, 1st it’s asking to choose 2, and then it’s saying minimum of 300 subnets (B & E are correct ) and max 50 hosts ( E is the closest to be correct but again it’s asking to select 2 answers )

    so 255.255.255.128 will give us 512 subnets and 126 hosts

    and 255.255.255.192 will give us 1024 subnets and 62 hosts

    if the Question was Max 300 subnets and min 50 host then B is the correct answer ( but again its stating select 2) go figure!

  10. Mark Terrell
    February 20th, 2013

    My guess is take the largest prefix (24) and subtract two or /22 and that would be Router A’s summary of the three networks; though /22 is not a choice.

  11. Mark Terrell
    February 20th, 2013

    Good questions gentlemen…keep em coming!

  12. abeer
    February 27th, 2013

    anyone can explain Q4 with another method i dont understand these solve

  13. Jah Rastafarai
    February 27th, 2013

    Passed CCNA, question 3 from here.

  14. usman
    March 1st, 2013

    hello can someone please explain me these question with explanation plz plzz i am waiting for reply…

    Q1 An employee of ABC company has moved to an office on a different floor. Although the
    administrator is able to telnet to all of the routers, the address of Switch F is needed in order to verify that the employee remains in the same VLAN. Which action could be used by the administrator to find the IP address of Switch F?
    A. Issue the show ip route command on Router A.
    B. Issue the show ip route command on Router C.
    C. Issue the show cdp neighbors command on Router B.
    D. Issue the show cdp neighbors detail command on Router C.
    E. Issue the show arp command on Router A.
    F. Issue the show arp command on Router B.

    Q2 Which of the following is a characteristic of full-duplex communication?
    A. It is a CSMA/CD network.
    B. It is a CSMA/CA network.
    C. It is point-to-point only.
    D. Hub communication is done via full duplex.

    Q3 A network administrator is configuring ACLs on a Cisco router, to allow IP access from the
    192.168.146.0/24, 192.168.147.0/24, 192.168.148.0/2,. and 192.168.149.0/24 networks only.
    Which two ACLs, when combined, should be used?

    A. access-list 10 permit ip 192.168.146.0 0.0.0.255
    B. access-list 10 permit ip 192.168.146.0 255 255.255.0
    C. access-list 10 permit ip 192.168.147.0 0.0.255 255
    D. access-list 10 permit ip 192.168.149.0 0.0.255.255.0
    E. access-list 10 permit ip 192.168.148.0 0.0.1.255
    F. access-list 10 permit ip 192.168.146.0 0.0.1.255

    Q4 A company has a small network,consisting of a single switch and a single router. The switch has been configured with two vlans,and route-on-a-stick is being configured on the router for inter-vlan routing. A trunk is configured to connect the switch to the router. What is the minimum number of router subinterfaces that are required for all the vlans to communicate?
    A. one
    B. three
    C. two
    D. zero

    plz answer meeee

  15. shshoor
    March 1st, 2013

    ques 3 was in the exam today

  16. MH
    March 4th, 2013

    @Usman
    Q2. C. (2 Dvcs connected P2P)
    Q4. C.(Eq 2 # of Vlans)

  17. jimmy
    March 6th, 2013

    Q5) ANS-C

    ANY ONE EXPLAIN PLZ

  18. Waqar
    March 8th, 2013

    Q3. E, F looks correct answer to me

  19. AN9
    March 9th, 2013

    HI THERE!
    I HAVE BEEN GIVEN THIS NETWORK 19.0.2.0/23 AND I HAVE BEEN ASKED WHICH OPTION IS VALID IP ADDRESS THAT CAN BE ASSIGNED TO A HOST?
    A) 192.0.2.0
    B) 192.0.2.255
    C)192.0.3.255
    D) 192.0.4.0
    I HAVE BEEN TOLD THAT THE ANSWER (B) BUT DO NOT UNDERSTAND HOW THEY GOT THIS.
    PLEASE CAN SOMEONE EXPLAIN TO ME HOW THEY GOT THIS ANSWER ?
    HERE’S MY EMAIL: an9one@gmail.com.
    MANNY THANKS IN ADVANCE

  20. mista-ug
    March 11th, 2013

    the subnet mask here will be 255.255.255.254
    so we have 256-254= 2subnets
    so, 0,2,4,6,8 etc are the sunets so from here you work your way out if you can get the broadcast address of each
    for example first network 192.0.2.0- 192.0.4.0
    the broadcast is 192.0.2.3 host address 192.0.2.255

  21. Question-05
    March 12th, 2013

    In Q5 Option D is also correct, What should we do to that…???

  22. Question-05
    March 12th, 2013

    @9tut
    In Question-5
    Option-D is also correct coz its in between
    172.16.1.128 subnet which ranges from
    172.16.1.128 to 172.16.1.254 so we can also assign 172.16.1.205 to any host, am i right..?

  23. daBeat
    March 15th, 2013

    @Question-5.. D is incorrect cuz it says Vlan 1… 172.16.1.128/25 is vlan 2

  24. El_Tarik
    March 16th, 2013

    @AN9
    192.0.2.0/23 borrows one bit from the 3rd octet, the last bit (1st octet = 8 bits, 2nd octet = 16 bits, 7 bits of 3rd octet = 23 bits total).
    The value of the bit is 2 (last bit is 1, second last 2, then 4, 8, 16, 32, 64 and 128).
    So we know that each network address (subnetwork address) will be in multiples of 2 starting at 0, so the first answer is incorrect as it is a network address.
    The third answer is 192.0.3.255. If you check the next network address it is 192.0.4.0. The broadcast address is always the next network (subnet) address minus 1, so in this case, 3.255 is the broadcast address so it can not be a host address.
    The 4th answer is 192.0.4.0 which is the address for the subnet 192.0.4.0 (multiples of 2 in the 3rd octet).
    The correct answer is the 2nd one, as it is 192.0.2.255. It is a trick question as when people see a 255 in an octet they assume it is a broadcast address but it is not so as this example proves. The next subnet address is 4.0 not 3.0, hence 2.255 is a valid host address.
    If you translate it all into binary, you will see this very clearly and easily. Let me know if you have more questions mate, cheers!

  25. Mahesh V
    March 17th, 2013

    Passed exam with 960/1000. THnaks 9tut.

  26. Anonymous
    March 20th, 2013

    @usman

    Q1 – D. Issue the “show cdp neighbors detail” command on Router C.
    this command with “detail” keyword will display all neighboring devices with their IP address.

    Q2 – C. It is point-to-point only.

    CSMA/CD is Ethernet protocol to detect collision in half-duplex. CSMA/CA is wireless protocol to avoid collision. Hub operate in half-duplex only.

    Q3 – E & F
    access-list 10 permit ip 192.168.148.0 0.0.1.255 will cover 192.168.148.0/24,. and 192.168.149.0/24 networks.
    access-list 10 permit ip 192.168.146.0 0.0.1.255 will cover 192.168.146.0/24, 192.168.147.0/24 network.

    Q4 – C. two

  27. Monika
    March 20th, 2013

    Yes, this is just I’m talking about!..Hotmail doesn’t roemve my email domain from black list.Michel, my apologies; I emailed you invoice from my Yahoo account just minute ago, please check your mailbox.Regards,Andrey.

  28. Modaser
    March 20th, 2013

    Wonderful and useful questions and answers !

  29. Charles
    March 21st, 2013

    Question 3:

    B & E are correct!!!.. to compy with the first condition… 300 subnets.. you have to figure out how many subnet bits u need to rob from the 3rd octet and pssibly from the 4th of the class B network scheme and how many are left for the host bits.
    Choose 2:

    B. 255.255.255.128 = 11111111.11111111.11111111.10000000 = /25 => 2^9 subnets = 512 (300 subnets) and 2^7 hosts = 128 -2 = 126 (50)

    E. 255.255.255.192 = 11111111.11111111.11111111.11000000 = /26 =>2^10 subnets = 1024 (300 subnets) and 2^6 hosts = 64 -2 = 62
    Hope this helps for everyone!!! Regards!!

  30. Yaser Arafat
    March 23rd, 2013

    A network administrator is configuring ACLs on a Cisco router, to allow IP access form the 192.168.146.0/24,192.168.147.0/24,192.168.148.0/2,. and 192.168.149.0/24 networks only. Which two ACLs, when combined, should be used.

    A.access-list 10 permit ip 192.168.146.0 0.0.0.255
    B.access-list 10 permit ip 192.168.146.0 255 255.255.0
    C.access-list 10 permit ip 192.168.147.0 0.0.255 255
    D.access-list 10 permit ip 192.168.149.0 0.0.255.255.0
    E.access-list 10 permit ip 192.168.148.0 0.0.1.255
    F.access-list 10 permit ip 192.168.146.0 0.0.1.255

    answer :E F
    can anybody explain to me this one? Thanks!

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  36. golu
    April 13th, 2013

    can anyone explain me Q.6

  37. Hes
    April 16th, 2013

    @golu
    Explanation is clear.
    Network 172.25.0.0 is a Class B which means sunbet should be 255.255.0.0 and third part would be changed, i mean 255.255.X.0 ( X part of network or subnet )

    Question say : divided into eight equal subnets which means : 255/8=32 and subnets will be : 0->32->64->96->128->160->196->224->256.

    ” ip subnet-zero ” command is configured which means you can use all subnets from 0 and third subnet would be :

    1. 0–>32
    2. 32–>64
    *** 3. 64–>96 ***

    It is Class B then in the network : 172.25.X.0 you need to use X part to calculate thefore third subnet would be :

    1. 172.25.0.0 –> 172.25.31.255
    2. 172.25.32.0 –> 172.25.63.255
    *** 3. 172.25.64.0 –> 172.25.95.255 ***

    As you look at the Answers : A C D are the correct.

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    April 28th, 2013

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  41. golu
    May 3rd, 2013

    @Hes thank you

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    May 4th, 2013

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  45. Kris
    May 13th, 2013

    In question 6, is it a practice to assign a .0 address to a host? Is this correct? I realize the host would be in the middle of the subnet, so it isn’t the network nor is it a broadcast. I just don’t recall everything seeing a host with a .0 address.

  46. Kris
    May 13th, 2013

    Oh, on question 6, guess I should question of the .255 answer as well…

  47. Rubie-sky
    May 26th, 2013

    Q6 .. you cant use the .255 address it’s for broadcasts .. the 0 likewise is the network ID, hence why you -2 from the amount of available hosts you have.

  48. Annoyn
    May 28th, 2013

    You can use the 94.255 address in Q6, because the broadcast of that subnet is 95.255 it’s a tricky one

  49. David Okeri
    June 5th, 2013

    Pliz i will sit for the CCNA exam next month, can someone send me the latest dumps on email: olesimbe@yahoo.com

  50. nicky
    June 17th, 2013

    I HAVE A QUESTION, FOR THE ONES WHO ALREADY TOOK THE EXAM, DOES THEY LET YOU TO USE ANY PENCIL OR PAPER, CU’Z TO SUBNETTING??
    I CAN NOT DO EVERYTHING WITH MY MIND :s

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