New CCNA – Subnetting
Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.
Question 1
Refer to the exhibit. Which subnet mask will place all hosts on Network B in the same subnet with the least amount of wasted addresses?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0
Answer: B
Explanation
310 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111 1110.0000 0000 -> 255.255.254.0
Question 2
Refer to the exhibit. All of the routers in the network are configured with the ip subnet-zero command. Which network addresses should be used for Link A and Network A? (Choose two)
A. Network A – 172.16.3.48/26
B. Network A – 172.16.3.128/25
C. Network A – 172.16.3.192/26
D. Link A – 172.16.3.0/30
E. Link A – 172.16.3.40/30
F. Link A – 172.16.3.112/30
Answer: B D
Explanation
Network A needs 120 hosts < 128 = 27 -> Need a subnet mask of 7 bit 0s -> “/25″.
Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.
Answer E “Link A – 172.16.3.40/30″ is not correct because this subnet belongs to MARKETING subnet (172.16.3.32/27).
Answer F “Link A – 172.16.3.112/30″ is not correct because this subnet belongs to ADMIN subnet (172.16.3.96/27).
Question 3
You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. Which network address and subnet mask meet this requirement?
A. 192.168.252.0 255.255.255.252
B. 192.168.252.8 255.255.255.248
C. 192.168.252.8 255.255.255.252
D. 192.168.252.16 255.255.255.240
E. 192.168.252.16 255.255.255.252
Answer: B
Question 4
Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?
A. 0.0.0.240
B. 255.255.255.252
C. 255.255.255.0
D. 255.255.255.224
E. 255.255.255.240
Answer: D
Explanation
A is not correct because it is a wildcard mask (not subnet mask).
This question is a bit unclear but we can suppose we have to begin with default subnet mask and “subnet” it. And the default subnet mask here should be class C: 255.255.255.0
For answer B: 252 = 1111 1100 -> with this subnet mask we can subnet up to 26 = 64 subnets but only 22 – 2 = 2 hosts per subnet -> B is not correct.
C is not correct because it is the default subnet mask of class C and that means we don’t “subnet” it.
For answer E: 240 = 1111 0000 -> There are 24 = 16 subnets but only 24 – 2 = 14 hosts per subnet < 26 hosts -> E does not satisfy the second requirement (of 26 hosts per subnet).
For answer D: 224 = 1110 0000 -> There are 23 = 8 subnets and 25 – 2 = 30 hosts > 26 hosts -> This is the correct answer.
Note: The number “5″ in ” with each LAN containing 5 to 26 hosts” is just used to trick you and it does not have any effect on our answer.
Question 5
An administrator must assign static IP addresses to the servers in a network. For network 192.168.20.24/29, the router is assigned the first usable host address while the sales server is given the last usable host address. Which of the following should be entered into the IP properties box for the sales server?
A. IP address: 192.168.20.14
Subnet Mask: 255.255.255.248
Default Gateway: 192.168.20.9
B. IP address: 192.168.20.254
Subnet Mask: 255.255.255.0
Default Gateway: 192.168.20.1
C. IP address: 192.168.20.30
Subnet Mask: 255.255.255.248
Default Gateway: 192.168.20.25
D. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway: 192.168.20.17
E. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway: 192.168.20.25
Answer: C
Question 6
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.
Question 7
You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each.Which IP address range meets these requirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
Answer: A
Explanation
Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bit 0s -> /27. Also the question requires the maximum number of subnets (which minimum the number of hosts-per-subnet) so /27 is the best choice -> A is correct.
Question 8
Which two benefits are provided by using a hierarchical addressing network addressing scheme? (Choose two)
A. reduces routing table entries
B. auto-negotiation of media rates
C. efficient utilization of MAC addresses
D. dedicated communications between devices
E. ease of management and troubleshooting
Answer: A E
Question 9
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
Explanation
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 (borrow 7 bits)
Question 10
Given an IP address 172.16.28.252 with a subnet mask of 255.255.240.0, what is the correct network address?
A. 172.16.16.0
B. 172.16.24.0
C. 172.16.0.0
D. 172.16.28.0
Answer: A
Explanation
Increment: 16 (of the third octet)
Network address: 172.16.16.0
-> A is correct.
Q9 sorry found it, the /30 is from the mask 252 = /30..
hi ! anyone help me how to clear the ccna exam.i’m going to write on march 10th… anyone have upated dumps mail it to vgangacse@gmail.com
plz help me question 3
Nadeesha,
You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. 3 Servers, so you neeed 3 IP address plus the subnet (+1) and broadcast address (+1) (you always always need 2 more) = 3+1+1= 5. You will need 3 bits (2^3 = 8) to cover this amount. So 256 – 8 = 248.
The only one answer with that mask is C. 192.168.252.8 255.255.255.248
can any one help me please see question number 304 from watson 314 question . i am no understanding it. can any one help me my email id is usman .anjum @gmail.com
Q2: Explain “E” is not belong to Marketing & “F” is not belong to ADMIN
hi why Q6 A?
I was really really having trouble to get to know subnetting. took me over 3 weeks but still couldn’t figure it out but then i found this guy on youtube and i am gleeful… I am sharing it with all the people who are having trouble doing subnetting. Keep me in your good thoughts.. here is the link http://www.danscourses.com/CCNA-1/subnetting.html
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Any time bro. I hope you got the concept. I know it gets so crazy and hard to get it but trust me one think you need to do is keep practicing. Only way it worked for me.
Why IP addressing scheme is called hierarchical? And what it meant by ease of managing and troubleshoot?
i can’t understand about Q4….anyone plz help to understanding abt Q4
Hi
Is this site recommended for someone who wants course material to study from scratch?
I only have Network plus and no other CCNA material, please advise?
Thanks.
Explanation Answer 9:
It’s much easier if you think like this.
You have point to point connections, so you know that you have 4 Adresses for each one. We have 113 links so 113 Subnets. 113Subnets x 4 Addresses = 452
We a subnetmask /24 you only have 256 Adresses so you know you have to take one more bit = /23. Than you’ll have 512 Adresses.
Why did 9tut delete my explementation off question 9???
it was so easy to understand like this-.-
*explanation sry
@Kayne: Your post were not been deleted, it is here: http://www.9tut.com/new-ccna-subnetting-2
Oh sorry. thank you
Q-7 ans should be B is correct
because 30 host means it comes in 32 block- 2^5 then 11100000
the set bit is to add with prefix length 10.188.31.0/23+3
the Ans is B.10.188.31.0/26
On Q.9. If the mask was /32, would the answer be B) ? Just getting my head around it.
On Q1 should be C the option corrrect not B
2^9 need 9 bits but the first bit that you start is 0
Q.5….its answer is d.why u select c..can u explain??
How to calculate subnet..its as easy as making tea…let take an example.we have a class c network.. IP 192.168.12.9…255.255.255.0…is it OK..well all u know its class c and the network bits are 24. …so write it like dis…192.168.12.9/24…got it.now on subnetting….if u want subnet u can add more network bits like..we already hv /24..now add 1 bit…now it is /25 ….now calculate that borrow bit in binary….2^7( why 7 ??…it is because the bit that we borrow was in the 7th position.. Got it),so 2^7=128…means we have 128 subnets if we use /25….
Now the host turn….from the above,,, 1 bit is borrowed…and how many were left…??? Answer is 7. [How??......we borrow 1 bit from last octet(8 bits)]………so 8-1=7…7 host bits means 2^7=128….128 hosts per segment…..
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Q4 – the question is about the englobing subnet or the individual subnets for 26 hosts?
I belive that the answer is the subnet TO BE subneted in 8 vlans of 26 hosts, and so the answer is 255.255.255.0 , 3 bits for subnets + 5 bits for hosts.
Please explain Q10 for me clearly.I don’t understand it.
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Q9: my (faster?) solution –
113 subnet with 4 “host” each (2 host+net.+brod.)
-> 113*4 = 452 total host
-> 452 2^9 = /23
;)
Q3 was in test today
First, you are asked to create 8 subnets with equal number of hosts with network 172.25.0.0. What you need to do first is divide 256 by 8 so that you’ll have 32 equal number of hosts.
256 / 8 = 32 -2 = 30
30 is use for valid hosts
1 IP Address use for Broadcast Address
1 IP Address use for Network Address
In total you have 32 IP Addresses in one subnetwork
Second “ip subnet-zero” it only means that the first subnet is used for the Router’s network interfaces which is First subnet: 172.25.0.0/19.
Last question, the third subnet which as you can see from the computation is
Third subnet: 172.25.64.0/19
If still confuse, contact me on my email: amariano1989@gmail.com
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Q7: I think the right answer should be B. Because we need 30 hosts +gateway address.
r u guys seeing the same ip addresses in the test for these subnetting questions?
please reply asap
thankyou
Q7: Correct answer should be B. The explanation that you need to borrow 3 bits for subnetting is correct, but your starting range is subnetted with a /23 (not a /24). So you are really adding 23+3 (not 24+3) to get the /26 subnet, not /27.
^^^ At least I thought that was the case. Everywhere I find this question online, they have the answer as A (/27), so maybe I’m wrong.
^^^ Q7 Yeah I’m totally wrong now that I think about it. The answer has nothing to do with the original subnet. Derp. Sorry!
Q7: if you choose /27, there will be only 29 addresses that can be assigned to hosts, because we need one address assigned to gaetway.
So, I think this is a trick question.
Q7: if you choose /27, there will be only 29 addresses that can be assigned to hosts, because we still need one address assigned to gateway of router, but /26 can meet the requirement of the network.
So, I think this is a trick question.
Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?
A. 0.0.0.240
B. 255.255.255.252
C. 255.255.255.0
D. 255.255.255.224
E. 255.255.255.240
Answer: D
For the life of me, I don’t understand why the answer is D (255.255.255.224 or /27). First off, /27 can give you 30 hosts at most but does it satisfy the question that you can subnet it for up to eight LANs with each LAN contatining 5 to 26 hosts? Of course not. The least subnet mask that answers the question is /24 (255.255.255.0) because you can borrow up to 3 bits (satisfies the first requirement of up to eight LANs) and ultimately ending up at /27 for the 5 to 26 hosts requirement.
Any help here? Every website I see say that the answer is D but I don’t understand why.
Question, I been studying these questions and practicing, are the questions on the exam the same or different, still having hard time getting it, and my exam is Friday
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Q7, say, plan to allow the maximum number of subnets with as many as 30 hosts each.
Maximum is /27 –> 30 hosts each
/26 is Minimun –> 62 hosts
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Q1 was in today exam
Question 9 was in the exam recently
detail of Q3…..?
What do you think about Q4?
If the network address range IS TO BE SUBNETTED for up to eight LANs, maybe in the host-ID i need 3 bits (for the lan) + 5 bits (for the hosts) with 8 bit at all… in this case the right answer could be 255.255.255.0.
@Alex
Your explanation to the question is absolutely logical. Thank you.
anyone can help me ??
many thx
host A address:192.168.1.78 255.255.255.224
host B address:192.168.1.130 255.255.255.192
which 3 answers are correct? and why?
A. host A IP address: 192.168.1.79
B. host A IP address: 192.168.1.64
C. host A default gateway: 192.168.1.78
D. host B IP address: 192.168.1.128
E. host B default gateway: 192.168.1.129
F. host B IP address: 192.168.1.190
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