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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
  1. Chris
    June 21st, 2020

    I think the words are mixed up on question three. The question states the network must have at least 300 subnets and the hosts must have a maximum 50 per network. Yet in the answer, you show that ” 2^7-2 = 126 > 50 hosts per subnet”..
    The answer shows 126 hosts is greater than 50 hosts. It can not be the correct answer if you have more hosts then the Maximum hosts per subnet. ???

  2. Dan
    February 22nd, 2022

    I agree with Chris. Can we get some clarification?

  3. A little cat
    March 26th, 2022

    Read the notice of the end Q3 pls Dan

  4. StefK
    April 3rd, 2023

    Link with questions not working , is made on Adobe Flash Player, i would suggest and update.

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