CCNA Subnetting Questions:

[Vijayb]

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.

[Vijayb]

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.

[Vijayb]

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.



(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.

[Vijayb]

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)



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.

[Vijayb]

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.

[Vijayb]

Question 7

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



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.

[Vijayb]

Question 1

Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?



A. 192.168.1.56/27
B. 192.168.1.64/26
C. 192.168.1.64/27
D. 192.168.1.56/26

Answer: B

Explanation

60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> "/26″. The question requires "wasting the fewest addresses" which means we have to allow only 62 hosts-per-subnet -> B is correct.

[Vijayb]

Question 2

Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANs in Phoenix but no additional subnets?



A. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

B. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

C. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

D. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

E. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

F. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

Answer: D

Explanation

All the above networks can be summarized to 10.0.0.0 network but the question requires to "represent the LANs in Phoenix but no additional subnets" so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to "move left" 2 bits of "/24″-> /22 is the best choice -> D is correct.

[Vijayb]

Question 3

Refer to the exhibit. What is the most appropriate summarization for these routes?



A. 10.0.0.0/21
B. 10.0.0.0/22
C. 10.0.0.0/23
D. 10.0.0.0/24

Answer: B

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.

[Vijayb]

Question 4

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= 128 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.



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.



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".

[Vijayb]

Question 5

Which address range efficiently summarizes the routing table of the addresses for router main?



A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21

Answer: C

Explanation

To summarize these networks efficiently we need to find out a network that "covers" from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (24 = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.

The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0... are different networks).

Therefore "move 4 bits to the left" of "/24″ will give us "/20″ -> C is the correct answer.

[Vijayb]

Question 6

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)



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

First we should notice that different VLANs must use different sub-networks. In this case Host A (172.16.1.126) and Host B (172.16.1.129) are in different VLANs and must use different sub-networks. Therefore the subnet mask in use here should be 255.255.255.128. In particular, it is 172.16.1.0/25 with 2 sub-networks:

+ Sub-network 1: 172.16.1.0 -> 172.16.1.127 (assigned to VLAN 1)
+ Sub-network 2: 172.16.1.128 -> 172.16.1.255 (assigned to VLAN 2)

-> B is correct.

The IP address 172.16.1.25, which is in the same sub-network with host A so it can be assigned to VLAN 1 -> C is correct.

To make different VLANs communicate with each other we can configure sub-interfaces (with a different IP address on each interface) on the LAN interface of the router -> F is correct.

[Vijayb]

Question 7

The network administrator needs to address seven LANs. RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used. What is the maximum number of usable IP addresses that can be supported on each LAN if the organization is using one class C address block?

A. 6
B. 8
C. 14
D. 16
E. 30
F. 32

Answer: E

Explanation

"The network administrator needs to address seven LANs" means we have 7 subnets < 8 = 23, so we need to borrow 3 bits from the host part (to create 8 subnets). We are using class C address block which has 8 bits 0 (the default subnet mask of class C is 255.255.255.0), so the number of bit 0 left is 8 – 3 = 5. Therefore the hosts per subnet will be 25 – 2 = 30 -> E is correct.

[Vijayb]

Question 8

Refer to the exhibit. What is the most efficient summarization that R1 can use to advertise its networks to R2?



A. 172.1.0.0/22

B. 172.1.0.0/21

C. 172.1.4.0/22

D. 172.1.4.0/24
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

E. 172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Answer: C

Explanation

Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24

Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (22=4).

[Vijayb]

Question 9


Gateway of last resort is not set
192.168.25.0/30 is subnetted, 4 subnets
D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1
C 192.168.15.4/30 is directly connected, Serial0/1
C 192.168.2.0/24 is directly connected, FastEthernet0/0

Which address and mask combination a summary of the routes learned by EIGRP?

A. 192.168.25.0 255.255.255.240
B. 192.168.25.16 255.255.255.252
C. 192.168.25.0 255.255.255.252
D. 192.168.25.28 255.255.255.240
E. 192.168.25.16 255.255.255.240
F. 192.168.25.28 255.255.255.240

Answer: E

Explanation

We have 4 routes learned by EIGRP:

D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1

These subnets are all /30 (as it says "192.168.25.0/30 is subnetted, 4 subnets". We have 4 successive subnets = 22 so we can go back 2 bits -> the summarized subnet mask is 30 – 2 = 28 and the summarized network is 192.168.25.16.