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ICND1 – Subnetting

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

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Premium Member: You can test your knowledge with these questions first via this link.

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

Question 1

[am4show have=’p2;’]What is the subnet address for the IP address 172.19.20.23/28?

A. 172.19.20.0
B. 172.19.20.15
C. 172.19.20.16
D. 172.19.20.20
E. 172.19.20.32

 

Answer: C[/am4show]

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

[am4show have=’p2;’]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

 

Answer: C[/am4show]

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

[am4show have=’p2;’]Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?

A. 192.168.1.8/29
B. 192.168.1.32/27
C. 192.168.1.40/29
D. 192.168.1.16/28
E. 192.168.1.48/29

 

Answer: C[/am4show]

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

[am4show have=’p2;’]Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)

A. 10.1.168.0
B. 10.1.176.1
C. 10.1.174.255
D. 10.1.160.255
E. 10.1.160.0
F. 10.1.175.255

 

Answer: A C D[/am4show]

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

[am4show have=’p2;’]Which one of the following IP addresses is the last valid host in the subnet using mask 255.255.255.224?

A. 192.168.2.63
B. 192.168.2.62
C. 192.168.2.61
D. 192.168.2.60
E. 192.168.2.32

 

Answer: B[/am4show]

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

[am4show have=’p2;’]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)

A. 192.168.4.61
B. 192.168.4.63
C. 192.168.4.67
D. 192.168.4.125
E. 192.168.4.128
F. 192.168.4.132

 

Answer: C D[/am4show]

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

[am4show have=’p2;’]An administrator must assign static IP addresses to the servers in a network. For network 192.168.20.24/29, the router is assigned the first usable host address while the sales server is given the last usable host address. Which of the following should be entered into the IP properties box for the sales server?

A. IP address: 192.168.20.14
Subnet Mask: 255.255.255.248
Default Gateway. 192.168.20.9

B. IP address: 192.168.20.254
Subnet Mask: 255.255.255.0
Default Gateway: 192.168.20.1

C. IP address: 192.168.20.30
Subnet Mask: 255.255.255.248
Default Gateway: 192.168.20.25

D. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway: 192.168.20.17

E. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway. 192.168.20.25

 

Answer: C[/am4show]

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

[am4show have=’p2;’]Given a Class C IP address subnetted with a /30 subnet mask, how many valid host IP addresses are available on each of the subnets?

A. 1
B. 2
C. 4
D. 8
E. 252
F. 254

 

Answer: B[/am4show]

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

[am4show have=’p2;’]Which two statements describe the IP address 10.16.3.65/23? (Choose two)

A. The subnet address is 10.16.3.0 255.255.254.0.
B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
E. The network is not subnetted.

 

Answer: B D[/am4show]

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

[am4show have=’p2;’]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[/am4show]

Explanation

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

 

Comments (8) Comments
Comment pages
1 2 308
  1. Natalie
    July 30th, 2017

    Dennis, the question is find two hosts in the same subnet, so in your subnet only one host available that’s why we should check next subnet which is 192.168.4.64 – 192.168.4.127, we can not use 192.168.4.64 as a valid host since it’s subnet address and we can not use the last valid host since it’s broadcast address 192.168.4.127, however answer C, D inside this subnet.

  2. Anonymous
    August 10th, 2017

    where do I find the questions please?

  3. Glen
    November 24th, 2017

    You can find the dumps on https://www.9tut.net/icnd1-100-105/new-icnd1-basic-questions-2 post from Irina

  4. Bob Dobbs Slack Master
    April 20th, 2018

    Question: How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.254.0?

    Answer: 128 subnets and 510 hosts

    Can someone explain the logic to this answer?

  5. Nelson
    May 1st, 2018

    @BDobbs –
    First off, take note that this is a Class B IP (last 2 octets usable).
    mask is 11111111.11111111.11111110.00000000
    a) host bits = (all the ‘0’s)= 9 bits
    therefore using formula (2^n)-2; 2^9=512 – 2 = 510 (minus 2 accounts for network and b’cast IPs).
    b) network bits = (all the ‘1’s in the 3rd octet) = 7 bits
    so, 2^n = 2^7 = 128 networks

    Keep playing with the numbers – you’ll eventually get the pattern!

  6. oya
    December 9th, 2018

    I need subnetting exercise

  7. Anonymous
    August 28th, 2019

    can someone explain why #2 isn’t -> B. 192.168.23.32 instead of C. 192.168.23.48

    if we increment by 16 -> 16 x 2 = 32; and 32 16 x 3 = 48; and 48 < 61

    but why isn't 32 right? Is it because 48 is a higher increment of 16 than 32? or something else?

  8. Anonymous
    November 9th, 2019

    Correct me if I’m wrong but for Question 2

    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

    The provided answer is C, which would be correct if the question was asking for Subnet Address.
    The question however asks for Network Address, and since it’s Class C IP address, I believe the correct answer should be A.
    Or am I missing something?

Comment pages
1 2 308