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.
Hello…!9tut,
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
Answer: D
My Ans. is B. (10.10.0.0/25 subnetted with mask 255.255.255.252)
Can you clear me? pl…
@koffy
subnet mask 255.255.254.0 = /23
Host addresses bits should not be all 1`s(broadcast address) and all 0`s(network address).
So:
1. 64.100.0000001{0.1111111} – not all 1`s – valid for a host
2. 172.55.0010000{1.11111111} – all 1`s – this is a broadcast
3. 64.104.1000000{1.00000000} – not all 1`s – valid for a host
4. 192.168.1010001{0.11111111} – not all 1`s – valid for a host
5. 10.165.55.255 – all 1`s – this is a broadcast add
6. 128.107.62.0 – all 0`s – this is a network add
@stantheman A+!
@koffy there are more ways to do subnetting, the answer is always the same.
@swati this question has been asked here before.
113 subnets of 4 IPs each. 113 * 4 = 452 IPs.
256 < 452 < 512. we will use a block size of 512 (2^9). this leaves 32-9=23 bits for subnetting.
@tuhin
you are mistaking MACs and subnets. please review your key concepts. thanks :)
if you want to improve your subnetting skills and the time it takes to answer such questions you should practice it some more.
there’s a subnetting question script that comes with any of my VCE downloads at this link:
http://www.tinyurl.com/xallaxvce
@Swati: We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 - 7 = 23.
So 10.10.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100
----------------------------------<7 bits-->—-
Thanks Stantheman. Did you purposely skip a step in your explanation?
Subnet mask: 255.255.254.0/23
And you said, host address bits should not be all 1′s (broadcast add), and 0′s (network add)
So basically the above subnet /23 is 1111111{0 00000000}….Is your analogy based on the fact that the last nine bits is all zeros, and when inverted becomes all ones. Hence, the host address cannot be all 1′s, and also 0′s? I mean pertaining this particular scenario. Please, explain. Thanks.
anonymous @koffy to stantheman…previous comment….
If given four routes with EIGRP configuration and auto summariztion disable.
Routes are 192.168.10.64/28….192.168.10.80/28….192.168.10.96/28….192.168.10.112/28. Which router command would be use to summarize these routes.
1. ip summary-address eigrp 1 192.168.10.64 255.255.255.192
2. ip area-range eigrp 1 192.168.10.80 255.255.255.224
3. summary-address 192.168.10.80 0.0.0.31.
4. ip summary-address eigrp 1 192.168.10.64 0.0.0.63
Answer is option 1.
Can any one share some light on this. Thanks.
@koffy
eigrp summaries are configured for each interface (that’s why the command starts with “ip…”)
.64/28 is .64~.79
.80/28 is .80~.95
.96/28 is .96~.111
.112/28 is .112~.127
per total we have from .64 to .127, 64 IPs.
a subnet to accommodate 64 IPs has a subnet mask of 255.255.255.192
answer: ip summary-address eigrp 192.168.10.64 255.255.255.192
remember: this is done under the interface configuration, we don’t use wildards here
@Anonymous
The concept of subnet mask is to identify what network an ip address belongs to
so: /23 is equal to 255.255.254.0 = 1111111.1111111.11111110.00000000
i didnt just inverted the remaining 9 bits to 1s or 0s.
in fact i didnt do such thing.
What i did was i converted the last 2 octet of the given ip addresses in to bits
Then applied the idea that if the hosts bits(the last 9 bits) are all 1`s then it is a broadcast address and it cant be used to a host computer.
Also if the hosts bits are all 0`s then it is a network address and cant be used to a host computer as well.
I hope i understood and answered your question..(–,)
Hello…! 9tut,
Thank you for clear me.
GIven the address 192.168.20.19.Which of the following are valid hosts on subnet ?
192.168.20.29
192.168.20.16
192.168.20.17
192.168.20.31
192.168.20.0
can someone help me in solving this question
@nishanth
you didn’t give the subnet mask. i’m guessing it was a /28 mask and 1st and 3rd answers could be correct.
there’s a subnetting practice script that comes along with the icnd1/icnd2/ccna VCEs here:
http://www.tinyurl.com/xallaxvce
@xallax thanks it was /28 ..But can you tell me how to solve it
@nishanth
/28 is 255.255.255.240
256-240=16 (this is the block size)
0~15
16~31
32~47
our subnet is between 16 and 31.
16 is the subnet ID, 31 is the broadcast address
hope this helps everyone with their subnetting
Subnetting:
CIDR- /9 /10 /11 /12 /13 /14 /15 /16
/17 /18 /19 /20 /21 /22 /23 /24
/25 /26 /27 /28 /29 /30 /31 /32
Subnet masks – 128 192 224 240 248 252 254 255
Block Sizes- 128 64 32 16 8 4 2 1
Q. How many hosts and how many subnets would you have with an address of 172.16.30.0/11
1. 254 subnets, 254 hosts each
2. 510 subnets, 126 hosts each
3. 2046 subnets, 30 hosts each
4. 8190 subnets, 6 hosts each
I’m leaning towards option 3, but the answer says option 2. (Could this be a mistake)
Can anyone share some light on this. Thanks.
@koffy
172.16.30.0/11 <– this is an invalid subnet mask.
172.16.0.0 is a class B IP, you can not subnet the first 2 bytes (octet=byte).
are you sure this is the right questions?
Yes, I’m certain of the question. Anyway, I did check back on the question. The only difference is the address 172.16.13.0/11. I must have mis-keyed in the 3rd octet. But yet still the first two octet remains the network address and cannot be subnetted. Right?
Thanks though…………
@koffy
if it were class A IP with a /11 mask then you would have had
3 bits for subnetting => 2^3 = 8 subnets
32-11=21 bits for hosts => 2^21-2 hosts (that’s 2 million IPs… minus 2!)
Thanks xallax…..
@koffy
i’m sorry, that’s the only answer i could give you
Your explanation is sufficiently enough. The concept is clear. Thanks again xallax.
Q. What is true about the following?
10.0.0.0 255.255.0.0 (Assuming ip subnet-zero is enabled)
1. 512 subnets, 65534 nodes per subnet, second subnet is 10.10.0.0
2. 256 subnets, 65534 nodes per subnet, second subnet is 10.10.0.0
3. 256 subnets, 65536 nodes per subnet, second subnet is 10.1.0.0
4. 256 subnets, 65534 nodes per subnet, second subnet is 10.1.0.0
5. 256 subnets, 65534 nodes per subnet, second subnet is 10.2.0.0
The answer is option 4.
xallax, would you be kind enough to break it down for me? or anyone. Thanks.
@koffy
10.0.0.0 255.255.0.0 (or /16)
1 octet for subnets: 2^8 = 128 subnets
2 octets for hosts: 2^16 – 2 = 65536 – 2 = 65534 hosts
subnets are:
10.0.0.0 ~ 10.0.255.255
10.1.0.0 ~ 10.1.255.255 <- second subnet here
correct answer: option 4
10.2.0.0 ~ 10.2.255.255
@koffy
10.0.0.0 255.255.0.0 (or /16)
1 octet for subnets: 2^8 = 128 subnets
2 octets for hosts: 2^16 – 2 = 65536 – 2 = 65534 hosts
subnets are:
10.0.0.0 ~ 10.0.255.255
10.1.0.0 ~ 10.1.255.255 <- second subnet here
10.2.0.0 ~ 10.2.255.255
correct answer: option 4
@koffy
it looks like you’re having difficulties with subnetting. there’s a script that generates subnet questions that comes with any of my VCEs here:
http://www.vce.comyr.com or http://www.ciscovce.com
i understand Question 3 until i saw the explanation, but i think it´s not a well developed and proposed
Thanks xallax. I will check that site.
I have a question not related to subnetting.
Q. JAX router with IP address of 192.168.11.1/24 is able to ping ATLANTA router with IP address of 192.168.11.2/24 successfully(using IP addresses). However, when the command telnet ATLANTA is entered from the JAX router the telnet connection failed. Which TWO reasons could be the cause of the failure.
1. The JAX router in not an entry in the host table of the ATLANTA router.
2. The JAX router does not have an entry for ATLANTA in its host table.
3. The host name command is not configured correctly on the ATLANTA router.
4. The host name command is not configured correctly on the JAX router.
5. Access to DNS server is not available.
The answer is option 2 and 5.
I’m not very clear with option 5. Can anyone explain or share some light on this. Thanks.
i understand Question 3 until i saw the explanation, but i think it´s not a well developed and proposed question
@koffy
the command is “JAX> telnet ATLANTA”
if the host is not defined on this device (option 2 applies) then the device will check its DNS (option 5 applies).
DNS resolves a name (“ATLANTA” is a name) to an IP.
Thanks for the help xallax…….
Hello all,
If I learn CBT Video training, 9tut.com and .net with exemcollection dumps, is it enought for CCNA exam ??
@fawad
read wendell odom’s book. read todd lammle’s book.
Thanks for the help xallax .i passed my exam ..
Hi all! can someone send me the latest CCNA actual exam?
please send it to bny521@yahoo.com
Appreciate anyone who will share it. Thanks!
@nishanth
congratulation and keep it up. go ccnp!
@ anyone can u send me the latest CCNA actual exam?
please send it to sab3001@hotmail.com
Appreciate anyone who will share it. Thanks!
Hello Guys,
Passed my exam yesterday with 933 scores, this dumps is almost 85% valid but the concept is very important because CISCO has replaced all the answers from the dumps.7 Simulation, 5 from VTP 9tut, one from EIGRP and ACL2 but the ACL was very difficult. Thanks for exam collection and 9tut.
Dear 9tut!
I need a help to understand this question: correct answer: B and D
Which two IP addresses could be assigned to a host with a subnet mask of 255.255.254.0?
A) 113.10.4.0
B) 186.54.3.0
C) 175.33.3.255
D) 26.35.2.255
E) 17.35.36.0
@ccna q&a
take a look at the subnet mask: 255.255.254.0
the network block ranges have this many IPs:
255.255.255.255-
255.255.254.000
_____________
000.000.001.255
first network starts at xxx.xxx.0.0, ends at xxx.xxx.1.255
second: x.x.2.0 ~ x.x.3.255
third: x.x.4.0 ~ x.x.5.255
rule: starts at a multiple of 2 in the 3rd bit, ends after 1.255 IPs
A) 113.10.4.0 is the subnet address, invalid IP
B) 186.54.3.0 is in the range 186.54.2.0~186.54.3.255. valid
C) 175.33.3.255 is the broadcast address for the 175.33.2.0 subnet. invalid IP
D) 26.35.2.255 is in the same range as B). valid
E) 17.35.36.0 is a multiple of 2.0 -> subnet address, invalid IP.
First of all we don’t care about 1st and 2nd octet, cuase that 255.255 and it could be anything, since interesting octet is 3rd octet, (254) and it increment by 2, subnet will be X.X.2.0 next subnet 4.0, 6.0 and so on till 254.0, and since 4th octet value is 0 – it’ll be 255, so available IP addresses range in our subnet 2.0 between x.x.2.1 – x.x.3.255
A) 113.10.4.0 – couldn’t be assigned to host since this is subnet number, we eliminate this
B) 186.54.3.0 – this IP falls in our range above x.x.2.1 – x.x.3.255, this is works for us
C) 175.33.3.255 – this is last IP and it’s broadcast address for our subnet 2.0 and couldn’t be assigned to host
D) 26.35.2.255 – !!!!!! this is trick, at first look this address seems to be broadcast IP, however by looking carefully and referring to our subnet IP ranges, this address still within our 2.0 subnet range, so this IP indeed works for us (this where was my hesitation, but it’s clear now after looking at details)
E) 17.35.36.0 – this is subnet IP and couldn’t be used
Thank you very much XallaX for your comment, indeed!
@ccna q&a
start from 0.0 or you will miss one subnetwork (x.x.0.0~x.x.1.255)
Xallax
I think that comes in place if question mentions about 0 – zero subnet, in this particular case there is nothing about 0 – subnet, so our subnet 2.0 and next subnet is 4.0 and so on…. correct me if I’m wrong
@ccna q&a
no, you are right, but you should always keep in mind there is that first subnet starting at 0
Xallax
ACK!
I do not understand how they came up with a subnet mask of /19?