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.
for through understanding study “binary to decimal” conversion. 3 ips are required from same sub-network. write on paper 128,64,32,16,8,4,2,1. Now write “1″ under each figure till 8. this means in last octet only 11111000.now add 128+64+32+16+8=248, if binary to decimal conversion is done it makes 11111000=248.
Hi all, can someone please explain the question 10, I calculate and I find that network 172.16.0.0 and 172.16.16.0 are true not only 172.16.16.0
thx
q3,q7,q8 ….Dec 18
for commenter below me that ask Q.10,
A is correct, B,C,D,E= wrong
subnet was /28 that means 0,16,32,64,..etc
and ip host : 172.16.28.252 so that between 172.16.16.0-172.16.31.255
Q5 today
pls ,explain q3, why choose (B).
For q3, if we calculate no. of usable hosts for 255.255.255.252, it comes out to be 2. We need to provide IP addresses for 3 servers. Thus options with 252 are eliminated. Now out of .248 and .240 , ,248 provides more number of subnets. Hence ‘B’. Please suggest if otherwise.
Can Someone explain Q7 please.
Can Someone Explain Q9….subnet mask
Q9.
Another way to understand this trick question. It says point-to-point links, ok, what do you need to assign ? Ip address, so you need 113 hosts instead of saying subnets. If you have to borrow 0s from the 1s, so you are getting hosts.
2^7=128, then you have to borrow 7 bits from the mask /30. You can do 30-7= 23 or /23. So thats finish.
thank you …
Q8 in my exam today . Passed
Q9 easy way.
/24 contains 256 addresses.
/23 contains 512 addresses.
Each /30 will consume 4 addresses (1 network, 2 valid addresses, 1 broadcast)
So, 256/4 = 64 (not enough)
512/4 = 128 (enough for 113 subnets)
my friends can i use calculator or paper in exam to solve subnetting questions???
Hi San,
As far as I know,, calculators and papers are not allowed
San,
You are given a laminated piece of paper to write on with a marker. You can do whatever you want on that during the test.
Hi SB and Ali
In Q3 , why not option d with 240?
i would go with “D” as this would give u most of the IPs which you cant have from the rest of the options, mask of 255.255.255.240 would mean bit mask of 28 which would mean 4 bits left for hosts that would bring you 2 to the power of 4 minus 2 = 14 hosts. Options a, c and e would get you 2 IPs and b would get u only 6 IPs using the same formula that we opted for d.
whats your opinion?
Hi friends, Nice site. How many lab questions come in exam
Q6 today. Praise allah for i have passed
Lubna,
.240 = 11110000; 2^4 = 16 subnets 2^4-2 = 14 host; it covers 3 host but max# of net. is 16
.248 = 11111000; 2^5 = 32 subnets 2^3-2 = 6 hosts; it covers 3 host but max# of net is 32
Question specifically asked the maximum number of subnets therefore .248 is correct.
Hope this helps.
Faizan,
For Q7; it asked for as many as 30 hosts which means we need 5 bits i.e 2^5-2 = 30 hosts
10.188.31.0/23
/23 = 255.255.254.0
9 bits available to play with X.XXXXXXXX
We need 5 bits reserved it from Right to Left(for Host) i.e 1.11100000 rest goes to network
so new Mask will be 255.255.255.224 i.e /27
Mask Above /27 have less then 30 hosts
Mask below /27 have more then 30 hosts which serve the purpose but it has less number of networks as we get with /27;
/26 11111111.11111111.11111111.11000000; i.e 62 hosts and networks 2^26 = this is sure have less # of net then 2^27
/27 11111111.11111111.11111111.11100000; i.e 30 hosts and networks 2^27 =
/28 11111111.11111111.11111111.11110000; i.e 14 hosts which is less then 30
THE QUESTION ASKED ” to allow the maximum number of subnets with as many as 30 hosts each.” From all available answers /27 is the correct options.
Hope this helps.
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
In my opinion answer is C and not D.
Question: :
Which subnet mask would be appropriate (X) to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?
Explanation:
26 Hosts : /27 ( 255.255.255.224)
In a /26 there is two /27
In a /25 there is four /27
In a /24 there is eight /27
So the answer is /24 = 255.255.255.0 => Answer C
Q1 .. Answer should be C not B
310 hosts 255.255.252.0
Q6 in my exam paper today got 1000 marks
Q1 is B
you have 310 hosts:
255.255.255.0 = can accomodate 254 hosts < 310
255.255.254.0 = can accomodate 510 hosts
255.255.252.0 = can accomodate 1022 hosts
since the question asked for the same subnet with least amount of wated address, so B is your best bet.
alaaah very easy questions
Q6 On 15th Feb.
i passed CNNA tody and Q6 was inclouded
q3,8,9
Q6. I guess all of the answer are incorrect, the summarization should be 172.16.32.0/19. Guys, what u came up with this question? Help???
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Passed my CCNA exam today (18th Mar)… Q8 in exam
Can you please sent the dumb to email michaelyoussef888@gmail.com thanks
could you pls justify ans for Q No 10 .
Please explain Q5 :(
Can you please sent the dumb to email biddut@live.com thanks
Passed the exam with flying colours 9tut does well
EIGRP 212
ACL MOD 3
VLAN Trunking
ACL
STP
Switching
Netflow
SNMP
Subnetting
Basically all the questions where here besides 4 new questions. You guys will be fine just practise these questions to get a good understanding of what CCNA is about. Make sure you understand what each section is about e.g. frame-relay, WAN, etc etc.
Because at the end of the day you can pass the exam but wont understand j@ck Sh!t about what networking is all about.
Can you please sent the dumb to email wasimcadd@gmail.com
thanks
Q4: hi anyone can explain me, I don’t understand below explanation …2^2-2=2..
For answer B: 252 = 1111 1100 -> with this subnet mask we can subnet up to 26 = 64 subnets but only 2^2 – 2 = 2 hosts per subnet -> B is not correct.
I think bit “1″ represent network. bit ’0′ represent host ..
to Martin:
You need at least 26 hosts.
However option B gives you this mask: 255.255.255.252
Written in binary: 11111111.11111111.11111111.11111100
There are only 2 zeros
2^2 = 4 – you may think you could address 4 hosts:
x.x.x.x00
x.x.x.x01
x.x.x.x10
x.x.x.x11
! HOWEVER !
IP address with all 0s is the NETWORK ADDRESS and can’t be used. Therefore x.x.x.x00 can’t be used to address host in LAN.
IP address with all 1s is called BROADCAST and also can’t be used. Therefore x.x.x.x11 can’t be used to address host in LAN.
THEREFORE, Everytime you are subnetting, you must subtract 2 (1 for network IP +1 for broadcast).
Therefore option B is not correct: 2^2 = 4-2 = ONLY 2 hosts. However you require 25 hosts.
2<25.
I hope, all is clear now.
Hi,
Can someone explain q6 please?
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Q2 : is the following explanation right ?
“Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.”
According to me, the ip subnet-zero refers to classfull networks and as 172.16 is a class B, its subnet-zero shoud be : 172.16.0.0 /n (n> or =16) and then 172.16.3.0/30 is not a subnet zero.
The no ip subnet-zero could have been set, it would not have changed the result.
Q4 is not cela
Q4 answer is not clear at all :
the answer should be 255.255.255.0 :
eight /27 subnets = one /24
It’s a classful /24 but what is the problem ?
Hey 9tut, thanks for all the material n stuff.
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