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 term VLSM stand for Variable Length Subnet Mask.
Here, we are dealing with Class B 172.16.0.0-172.31.255.255…………../16
First 2 octet is the network, and last 2 octet is the host. This means, we can only subnet the last 2 octet.
Options A, B, C, E, all are in the Class B with different mask. The router will run a search before summarizing. The router will choose the highest mask /16. In the eyes of the router, it will be able to subnet /17, /20, /24.( Summarizing all of these into one /16)
By the same token, if /17 is the highest mask in the options you provided, then the router will chose the /17.
VLSM can be sometimes confusing. The more you practice, the better.
I hope this helps. Thanx.
Plz solve these Questions in Detail
1. How can I get 4 Networks and every Network has only 25 valid IP. Class is C-192.168.1.0 ?
2. How can I get 6 Networks and every Network has only 2 valid IP. Class is C-192.168.1.0 ?
3. How can I get 2 Networks and every Network has only 14 valid IP. Class is C-192.168.1.0 ?
plz send me all the answers on my mailing address as soon as possible :- manish_730@yahoo.com
The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the valid IP address of this host?
A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14
@innocent
192.168.5.33 /28 means that this IP is part of
192.168.5.32 ~ 192.168.5.47
the only valid option was C, 192.168.5.40.
read up 9tut’s subnetting tutorial here:
http://www.9tut.com/subnetting-tutorial
Hy guys,
Can you please email me the latest dumps on rafan.fatima@gmail.com
hi everybody, i wanted to wish everyone good luck . I’ll try to pass my exam next week and i was just wondering what is the proportion of questions about subnetting during the exam. Thank you.
plz help me out in Q.5 of option c. .
Please someone send me the “Latest CCNA Dumps” on my mail:-
mailtorajusharma@gmail.com.
Help Someone…..And one day you will be helped by Someone”
Because “Every action has an equal and opposite reaction”
Hi Guyzzz
can somebody suggest me a good website which is somwhat descriptive than 9tut.come… i am looking for the step by step solution of the problems.. also please email me the latest dumps on libenambooken@gmail.com
@xallax
cant we do summarization using 172.16.0.0 / 17 as well for Q7?
@haciii
/17 would be
172.16.0.0 ~ 172.16.127.255
our summary address has to go all the way to 172.16.179.255
172.16.0.0/16 is the only correct option
@xallax
Q6. 8 subnets mean we need four bits for 8, so we have to add four 1 bits in 3rd oktet, hence increment becomes 2^4 = 16, you are using the formula 256 / subnet for the increment but I actually don’t get it.
For 16 subnet for ex. it gives 16 inc. but we need 5 bits for 16. hence adding three 1 bits gives us 32 increment. Am i False
@anonymous
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?
8 subnets; subnet-zero is usable: that means 2^3 bits for subnets
our subnet mask would be:
8+8+3+0 bits
255.255.224.0
increment is 256-(value of last bit, 224) = 256-224=32.0
subnets are:
172.16.0.0 ~ 172.16.31.255
172.16.32.0 ~ 172.16.63.255
172.16.64.0 ~ 172.16.95.255
…..
add 32.0
…..
172.16.224.0 ~ 172.16.255.255
we’re only interested in the 3rd one: 172.16.64.0 ~ 172.16.95.255
A – 172.25.78.243
is
B – 172.25.98.16
isnt
C – 172.25.72.0
is
D – 172.25.94.255
is
E – 172.25.96.17
isnt
F. 172.25.100.16
isnt
—–
pay attention to this: how many subnets? is subnet-zero usable? how many hosts per subnet? which address should i assign to which device?
who has the latest dump, keebin26@hotmail.com . thank you
the ip supnet zero command is not configured on a router .
what would be the ip adders of Ethernet 0/0 using the first
available address from the sixth subnet of the network 192.168.8.0 /29
A- 192.168.8.25
b- 192.168.8.41
c- 192.168.8.49
d- 192.168.8.113
@AS
/29 allowed for 8 hosts…..2^3
Range>>192.168.8.8~192.168.8.48……..First to the sixth
Therefore using the first available address from the the sixth subnet, mathematically is going to be:
8*6=48…………192.168.8.49(subnet zero not configured)…..Option C is your answer.
can anyone one explain me answers to Question 1. I understand how you got subnet address and broadcast address but if you look at subnet mask 255.255.255.224 and try to look valid host for option 1 you should be incrementing second octent so your second octet should increase in increment of 32,64,96,128…. so on while you are simply looking for these values in last octent which i feel is incorrect. Pls explain as i am missing a huge thing here
@satish
subnet mask of 255.255.255.224
increment of 256-(value of last incomplete byte – 224) = 256-224 = 32
valid
A – 15.234.118.63
this is the broadcast address for x.x.x.32~x.x.x.63
B – 92.11.178.93
valid address on x.x.x.64 ~ x.x.x.95
C – 134.178.18.56
x.x.x.32 ~ x.x.x.63, valid address
D – 192.168.16.87
same as B
E – 201.45.116.159
broadcast address on x.x.x.128 ~ x.x.x.159
that 159 at the end is 32*5-1, that’s how you can tell fast it’s a broadcast address
F – 217.63.12.192
network id, 192 is 32*6
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Hi guys,
My sincere appreciation for this site as it really helps. I am new in the the forum and intends to write my ccna soon. My question is. Is there a quicker way to calculate the no. of subnets/host in an exams environment due to time factor without having to do the binaries?
Thanks
@Ngoroko: Yes, there is a way to calculate subnets/hosts without converting into binary, please read http://www.9tut.com/subnetting-tutorial
@9tut
by default
ip subnet-zero this command is enable or disable in router?
by default it is enabled on cisco routers.
good (free) subnetting video tutorial: http://www.bosscbt.tv/uncategorized/intro-to-ip-addressing-part-i/
@BFromA
Thank you…
yes but not
I thought i was doing alright with subnetting, i did subnetting questions 3 and passed got them all right. I now did this one and only got 3 right??? Back to the drawing board, any advice people of what specific part of subnetting to cover?
about question “the ip subnet zero is not configured on router …” .I think answer is B so “the ip subnet zero is not configured ” mean ip subnet zero (defaul) on router -> sixth subnet is 192.168.8.40 -> first available address is 192.168.8.41
Hi guys. I’m just wondering,
are these two hosts on the same network? do they communicate within the LAN without a router?
PCA: 192.168.1.1/24
PCB: 192.168.1.2/25
@tohritz
yes, they will communicate even tho the subnet mask is not the same.
PCA is on subnet 192.168.1.0 ~ .255
PCB is on subnet 192.168.1.0 ~ .127
they are both on the subnet of .0 ~ .127 so they can “see” each other
@xallax
thanks for the clarification
Am having trouble with solving a summarized subnet.!!!!…..
can any one help please!!!!
Need help on this question
You are designing a subnet mask for the 192.168.211.0 network. You want 50 subnets with up to 1 hosts on each subnet. What subnet mask should you use?
answer is 255.255.255.252 how?
@peter
because that’s the biggest subnet mask you could ever use to subnet an IPv4 network
a subnet mask of 255.255.255.252 would give you 4 IPs per each subnet. 1 for the subnet, 1 for broadcast, 2 for hosts.
A little help, please
Confused on Question number 5 above:
Correct Answer are: B C F
My choice would be: B D F
How come C is the correct answer instead of D?
Thanx xallax
Could someone please email the latest dumps, my email is alljunkmail74@yahoo.com.au. Thank you.
explain briefly subnetting and supernetting
@Aonie:
VLAN1 – 172.16.1.0/25
VLAN2 – 172.16.1.128/25
205 > 128 so host 172.16.1.205 in VLAN2.
Hi 9tut… Hi Guys! Can you please help me… I will take exam this Feb. Please send me latest dump so that I will have an idea for the exam.. rico.blake@ymail.com
Thanks Guys!
Hi, am taking an exam on the 27 Feb can you please email me the latest dumps, my email address is ntaopajm@telkom.co.za
Regards
Assuming you can use subnet zero and all-ones subnet.
How many bits must you borrow to create 4 subnets?
A) 1
B) 2
C) 3
D) 4
E) 5
@basha
B. 2 bits
Hi all am taking an exam on the 17 Feb.can you please email me the latest dumps, my email is d_danial1905@hotmail.com
Regards
Hi all am taking an exam on the 20 Feb.can you please email me the latest dumps, my email is
mrashidjut@gmail.com
Hi every one, I am taking my CCNA exam on the 25 Feb. Can you please send me the latest dumps, my email is dur_dip@yahoo.com
How to download this files plz suggest me.
vinodrawal94@yahoo.in
Can you guys take a look at this question and confirm if I’m right or wrong? By my calculations neither of the answers to the questions are correct because the mask creates 16 usable subnets with 14 hosts per subnet, yet that is not an option from either of the answers.
1. Which of the following statements would best describe following the IP address and subnet?
192.192.192.56
255.255.255.240
This is a Class B address that is masking 12 bits for subnetting, yielding 14 valid subnets.
The subnet mask is borrowing 4 host bits which creates 14 usable subnets with 14 usable hosts per subnet.
This is a Class C address that is using 4 host bits which creates 16 usable subnets with 16 usable hosts per subnet.
This is a Class C address that is using 4 host bits which creates 14 subnets with 14 hosts per subnet.
@kevin
yes u r right, there is no option that best suites the answer. for the mask of 255.255.255.240 u will have 4 host bits, 16 subnets, 14 usable hosts per subnet………
Excuse me, I need help:
The question is : refer to this routing table which rout will Router take to send packet with IP 10.1.5.65?
10.1.1.0/24 e0 directly connected
10.1.2.0/24 e1 directly connected
10.1.3.0/25 s0 directly connected
10.1.4.0/24 s1 directly connected
D 10.1.5.0/24 e0 10.1.1.2
D 10.1.5.64/28 e1 10.1.2.2
D 10.1.5.64/29 s0 10.1.3.3
D 10.1.5.64/27 s1 10.1.4.4