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.
Yes Nicky, They usually provide you with a pen and a pad of paper for the test. but restrict any outside materials that you might bring with you in case you have notes jotted in the back or other underhanded schemes people try to cheat on the tests (not that you would but some people do try)
@nicky they’ll only provide you w/ a erasble marker and plastic paper/board. no other materials used from outside.
Please someone describe question-4. I think D is correct.
Rubel,, Letter “D” is belong to the first network w/c is 115.64.0.0 – 115.64.3.255/22.
what are the easy ways to tackle with subnetting questions in class B.those guys who have mastered subnetting.
hey guys plz tell me the ip addresses are same in exam? or not?
please can some one show me websites that i can read on subnetting basics to get the clear understanding.
Is it possible to use a network of 192.168.0.0 and have subnets 192.168.64.0 and 192.168.65.0 for one vlan and use S/M 255.255.254.0? Thank you for your response.
@ ibrahim
July 15th, 2013
find on youtube to understand subnetting
FREE URDU CCNA Lecture 11 Subnetting
Will they let you use a calculator?
Question 6??? – It’s obvious how “32″ came about dividing the network in 8 equal parts, however I do not understand how /19 is the corresponding subnet mask?
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@drtedd
Since your default mask is /16 and you need 8 networks from the provided network you need to borrow however many bits will allow for 8 networks.
So…
2^1 = 1
2^2 = 4
2^3 = 8. There are your 8 networks. You already have a /16 and you borrowed 3 bits which now gives you /19 16+3.
@Ruth, let me try to understand your question: you mean you’ve got 2 subnets:
192.168.64.0, and 192.168.65.0 and subnet of 255.255.254.0, given these 2 subnets, there’s only one posiible subnet mask we can work with and that is 255.255.255.0, and this is how we find our increment with this subnet mask: 256-256=1 and we will start adding 1 up to 255, but as you mention subnet mask 255.255.254.0, this mask gives us an increment of 2, which means our subnets will start, if subnet zero is being used from: 192.168.0.0, 192.168.2.0,……192.168.64.0, 192.168.66.0, we should use the increment(+2) in the third octet, as we are working with this subnet mask, that’s the way i try to understand your question, i hope it helps….
@dominic achira, the easy way to tackle with subnetting in class B is to find out first the subnet and the broadcast the address in which the address is for example, given the address 172.16.66.10/18, what subnet and broadcast is this address on? we all know that the default mask for a class B is a /16, then here we have 18 (18-16=2), 2 bits have been borrowed and we put them in the third octet and it will sit on 64 and 64 is our increment, we are working in the third octet, from our example 66 is between 0 and 64 and the next subnet is 128 since our increment is 64 and to find the broadcast for subnet 64 we take the next subnet which is 128-1 and we have 127, and the host addresses are any numbers between the subnet and broadcast, the most important thing to keep in mind is to know which octet you are working, it becomes easy once yoj know the number of bits borrowed,….my small contribution, i hope it makes sens to you
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An exam question read as follows:
You are required to provide IP addresses by subnetting the following network, 172.16.0.0/16.
The addresses are needed for the company’s 4 departments, the server room and for point to point WAN links.
The requirements of the network are as follow:
1) VF- 1000 host addresses
2) HR – 350 host addresses
3) FIN – 100 host addresses
4) LN – 50 host addresses
5) SERVER ROOM -20 host addresses
6) LINK 1,2,3,4 and 5 – 2 host addresses (these are point to point links, such as between 2 routers)
For EACH network show
a) How many bit/s did you use to provide the needed IP addresses? [6]
b) How many bits remained and how many networks would these bits create?[12]
c) What is the new subnet mask? [12]
d) For each network, generate the first and last valid IP addresses, and the broadcast addresses.(Do this in a table) [50]
Hey guys I got a question on number 6 . check the following
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).
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hi,
just want to ask if are the subnetting questions here the same with the actual ccna exam?
please reply. thanks.
Hi everyone!
Can anyone explain where did
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?
According to sekhar
Ans 14
Explanation
RIP version 1 does not support VLSM information, so all networks must have the same subnet mask. In the network above, there are a total of 12 networks (6 LANs and 6 different point to point WAN connections). Therefore, if each of the 12 networks use the 255.255.255.240 subnet mask, there will be a total of 16 networks with 14 usable hosts on each LAN.
Can anyone explain where the 6LANs and 6 WAN connection come from?
Please Im a bit confuse..
Thanks in advance
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hi, not really understand explanation for Q1. Anyone?
@ Nick Ques 1 Mask 224 gives block size of 32. You need to know this.
So for A – 15.234.118.63 The networks are 118.0 118.32 118.64 118.96 etc up to 224 and the broadcast are 118.31, 118.63, 118.95 etc Its explained there below the question. You cannot use 118.63 as a host cuz it is a broadcast address. Let’s try E…. E – 201.45.116.159 Networks are 116.128 116.160 116.192 up to 224….. (116.160 -1 ) = broadcast address which gives you 116.159. You cannot use E for a host. Its a broadcast address. Hope that helps but 9tut explanation is great.
Q. 4 the answer B. 115.64.7.64, would it be the broadcast network ID based on the magic #4, network 0,4,8,12,16 etc, so 7 would be the broadcast IP. Please reply have exam tomorrow.
Atif,
Usable network addresses for the 115.64.4.0/22 network is
115.64.4.1 to 115.64.7.254
115.64.4.0 – is the network address
115.64.7.255 – is the broadcast address
7 is the last number of the 3 rd Octect
But the host part is in the forth octect can still change from 0 – 254 while 3rd octect is 7
Good Luck for your exam !
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do they let using calculators ?
A national retail chain need to deisgn 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 ?
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Question 6)The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 please i need explanation on this.
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Netstack College has received a grant to wire all of its six buildings for 10/100 Ethernet. Four of the buildings contain classrooms and laboratories and will need to support 250 network connections each. One building is for administration and will need to support 180 network connections. The Administration Building also houses the Computer Center where the college servers and network backbone are located. The sixth building is the library, which must support 80 network connections.
You have been directed not to use a flat network of 1260 IP addresses assigned from a Class B network range. Instead, he would like to use multiple subnets.
Playing the role of the network administrator, what steps would you take to accomplish this setup using subnets instead of using a flat network of 1260 IP addresses assigned from a Class B network range? Determine and list the IP address ranges and masks associated with each building and explain the process you used to obtain the IP addresses and masks showing the obtained addresses and masks.
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General question about subnetting… what is the usual average time to solve these type of questions?
Can anybody tell that in Q.5 The switch used is Intelligent(L3) or not? Since it divides the Broadcast Domain as well.
I ask the Administrator as well.