CCNA – Subnetting
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.
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”.
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.
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.
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?
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.
@Anonymous
the answer is A and D
@smart Technology, A and D are the best answers. thanks
Plz dont charge for tut…
I really like your way for Q7…you are brilliant, thank 9tut so much….
@9tut
tnx for q7 explanation……a really nice explanation now i got summarization…….
@smart technolgy, @anonymous, @sniffer
A is ok, but how D is the broadcast address in the q asked by Anonymous..
plz rply… i hv my xm in few days……….
9tut thank you for explaining all these questions. You can find brain dumps anywhere but only you explain the, “why” nice and easy. Even after reading the books I didnt understand some of these, but you get right to the point. I am lucky to have found this site. thanks again…usa
AnkushK, the /30 means there 4 bits in each network. i.e. 172.23.2.0 is a network address and the 172.23.2.3 is its broadcast address. Knowing this you can find the closest network address to D. is 172.23.2.200 (count up by 4′s or just realize 200 is a multiple of 4). Therefore, you can see that .203 is its broadcast address.
Actually it is off by 2 the formula is 2(n-32)-2 thats 2 to the nth power minus 32 which gives you the nuebmrs in the chart and then you subtract 2 no gateway is not a reserved address.Thanks though Jay!
This is just great This is just great stuff. For someone who wants to get into the netowrking field, like Cisco or somthing else, this is very good for them. It’s well animated and provides very good visual help. Great job.Oh, I am taking a Cisco Academy Class if you were wondering.
@9tut
What are the usable host on the 96.6.128.0/19 network?
A. 96.6.128.1 through 96.6.143.254
B. 96.6.128.128 through 96.6.134.254
C. 96.6.128.0 through 96.6.134.254
D. 96.6.128.1 through 96.6.159.254
E. 96.6.128.1 through 96.6.135.254
answer: E
can you please explaine for me how they got the answer?
From my understanding I got D. Where did you get E from? From increments of 32 that come from a /19 subnet it would be 192.168.128.0 – 192.168.159.255, then minus the first and last address….
oops. Meant 96.6.128.1 through 96.6.135.254
I love you 9tut :))
I posted a subnetting video on Youtube, “Easy Subnetting Without Math”:
http://www.youtube.com/watch?v=_S_zJG1f8bE&feature=plcp
The video and my manual might be useful to some people who have difficulty with subnetting, although the video is boring.
If I have the following example below. What is the quickest way (from anyone’s experience), to find the network if I already know that I have block increments of 8?
86.248.259.79/29
Network: 86.248.259.?
Subnet: 255.255.255.248
Hi, I am sorry but I think that on question 3 there is only one correct answer and it is D.
B and E are worng becuase on both HOSt IP are off the limit as the question says that they need a Max of 50 IP addresses.
Could you please take a look and let me know if Im right?
Thanks guys.
@ Q7
if there is an option for 172.16.0.0/17
then what would be the right answer?
172.16.0.0/17 or 172.16.0.0/16
IU agree the question 3 answer is backwards. the answer should be .224
Q1: What are valid hosts on the 3rd subnet of 172.16.0.0/20 [ Assume ip subnet 0 is configured ]
Select 3 choice
A) 172.16.31.250
B) 172.16.45.23
C) 172.16.176.34
D) 172.16.32.35
E) 172.16.0.125
F) 172.16.39.178
G) 172.16.5.78
I am sure that Q6 wrong answer because , It mention in question that the ip subnet-zero command is configured on the router , so you have to ignore the zero subnet since it’s been used already on the router so the third subnet will be 172.25.96.0 so the answers should be:
B – 172.25.98.16
E – 172.25.96.17
F – 172.25.100.16
i am very sure about this because why they wrote in the question about the subnet zero in first place why they mention it ?? it’s a tricky question.
@alaa
B,D,F
Quote ”
I am sure that Q6 wrong answer because , It mention in question that the ip subnet-zero command is configured on the router , so you have to ignore the zero subnet since it’s been used already on the router so the third subnet will be 172.25.96.0 so the answers should be:
B – 172.25.98.16
E – 172.25.96.17
F – 172.25.100.16
i am very sure about this because why they wrote in the question about the subnet zero in first place why they mention it ?? it’s a tricky question. ”
The reason the subnet zero is mentioned is so u count it as the 1st subnet and not skip it and think that 172.25.32.0 /19 is the 1st . therefore :
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
is correct range is 65-95
Hi guys can u send me a reviewer regarding ccna co’z im gonna take the exam within this month plzzzz. oehinabio@gmail.com
Q6
I agree with Soulust :
If we divide the address 172.25.0.0 in 8 subnets, the resulting subnets will be
1. 172.25.0.0
2. 172.25.32.0
3. 172.25.64.0 This is the third subnet
4. 172.25.96.0
5. 172.25.128.0
6. 172.25.160.0
7. 172.25.192.0
8. 172.25.224.0
and, the ” ip subnet-zero” command tells the router that those subnets are NOT reserved.
Addresses that fall in the 3rd subnet will be from 172.25.64.0 —- 172.25.95.255
A, C and D are in this network range :
A – 172.25.78.243
C – 172.25.72.0
D – 172.25.94.255
(“those subnet” = first and last ones !)
9tut,
On Q5, could you breakdown how the IP address 172.16.1.25 is correct tobe assigned as host in VLAN1 with a /25 mask thus incrementing by 128 … which is how I understand it. Could you please explain this to me a bit more …. the other aspects of the question I understand. Thx
You already understood that 255.255.255.128 was the netmask.
So, let’s find the network address for VLAN1 , concentrate on the last octet :
we know host address .126 : 01111110
we know the mask .128 : 10000000
So the network address is : 172.16.1.0
The usable ip range is from 172.16.1.1 to 172.16.1.126.
172.16.1.25 is in this range and can be assigned to VLAN1
Sorry for my bad english o/ (french is native language).
Hi, every1.
Pls, help me to understand why question 4, answer F
F-115.64.12.128
Is wrong.
4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100
Never mind, I got it!
Can somebody plz explain this question…
The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme best defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses
A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252
i dont get why B is correct, E makes more sense to me since you are subnetting from a Class A address.
TY 9tut.
Today I have passed the CCNA. (860/825)
50 questions 3 labs (VTP, EIGRP, ACL). 35 from 9tut.
Also thanks a lot Brar and Sekhar (still valid from examcollection)
Ty again 9tut
from here none
Q5. Can anyone please explain to me, how we know the addresses of vlans are class B?
@reader
notice that both VLANs share the same /24 subnetwork (172.16.1.0/24) and it got divided into 172.16.1.0/25 and 172.16.1.128/25. the subnet mask in use is 255.255.255.128 on both subnets.
B is correct
hosts in VLAN1 have IPs in the range of 172.16.1.1~172.16.1.126 (judging by the subnet mask we figured out above and the info provided by the diagram – the IP of a host).
C is correct
the diagram shows only 1 connection from the switch to the router. the switch is also connected to 2 VLANs. the router is needed to route traffic between VLANs.
1) the link between the switch and the router must be a trunk link to allow frames from multiple VLANs to pass.
2) the router’s physical interface must be ‘split’ into subinterfaces (logical interfaces) and set to the appropriate encapsulation type (most likely dot1q)
F is correct
to better understand subnetting (which is essential for your cisco career) please visit:
http://www.9tut.com/subnetting-tutorial
!!!!!!!!!!!!!!!!!!!Great Work 9tut!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
UthmanHassan, Good explainations. Thanks
I finally pass the exam, all the stuff here are great study material.
Question 7. should be /20 and not /16 some please explain
i have issues with Q5 too:
Class B /25 = 126 hosts.
the vlans have 114 and 84 hosts, thats way more than 126
2x2x2x2x2x2x2-2 = 126
Class B /24 makes more sense.
256 subnets and 254 hosts per subnet
can someone pliz explain how we got /19 on question 6? thank you
What are the usable host on the 96.6.128.0/19 network?
A. 96.6.128.1 through 96.6.143.254
B. 96.6.128.128 through 96.6.134.254
C. 96.6.128.0 through 96.6.134.254
D. 96.6.128.1 through 96.6.159.254
E. 96.6.128.1 through 96.6.135.254
answer: E
can you please explaine for me how they got the answer?
@Hort..according to my calculation it should be D.Not sure how E can be the answer. The same question has been asked by one gentleman above as well. /19 is 224 which means 256-224= 32. So first IP is 96.6.128.1 through 96.6.159.254. E doesn’t add up. I would be very glad to know how E can be the answer here.
I am taking up CCNA next week.
Please send me the latest dumps Brar ans shekar
shyona.08@gmail.com
Hi All,
Looking to sit CCNA exam in a couple of weeks, but can’t seem to grasp subnetting. Has any1 got any tips or tables that I could use/memorize to take to the exam with me.
Thanks
Can anyone suggest how to convert from VCE to PDF..
i am unable to install the VCE ..it s not working properly..
Question 3 — I think the correct answer should be D – 255.255.255.224.
The question states there is a maximum of 50 hosts per subnet
Can anyone agree or disagree?
Nevermind, in the explanation it mentions that the wording is incorrect.
I disagree with you Rob, B and E are correct. you get 9 bits to get into 300 subnet which is 255.255.11111111.10000000=255.255.255.128 and 6 bits to get into 50 but this time around you save the hosts which is 255.255.255.11000000=255.255.255.192
I agree with you renolph B & E are correct.
from the question we must take consider both networks & hosts having that in mind the answers were right