CCNA – Subnetting Questions 4
Here you will find answers to Subnetting Questions – Part 4
Note: If you are not sure about Subnetting, please read my Subnetting tutorial.
Question 1
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 2
Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANs in Phoenix but no additional subnets?
A. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
B. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
C. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
D. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
E. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
F. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
Answer: D
Explanation
All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is correct.
Question 3
Which address range efficiently summarizes the routing table of the addresses for router main?
A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21
Answer: C
Explanation
To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (24 = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.
The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0… are different networks).
Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.
Question 4
Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?
A. 192.168.1.56/27
B. 192.168.1.64/26
C. 192.168.1.64/27
D. 192.168.1.56/26
Answer: B
Explanation
60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires “wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is correct.
Question 5
The network technician is planning to use the 255.255.255.224 subnet mask on the network. Which three valid IP addresses can the technician use for the hosts? (Choose three)
A. 172.22.243.127
B. 172.22.243.191
C. 172.22.243.190
D. 10.16.33.98
E. 10.17.64.34
F. 192.168.1.160
Answer: C D E
Explanation
From the subnet mask of 255.255.255.224 we learn:
Increment: 32
Network address: In the form of x.x.x.(0,32, 64, 96, 128, 160, 192, 224)
Broadcast address: In the form of x.x.x.(31,63,95,127,159,191,223)
-> All IP addresses not in the above forms are usable for host -> C D E are correct answers.
Question 6
In the implementation of VLSM techniques on a network using a single Class C IP address, which subnet mask is the most efficient for point-to-point serial links?
A. 255.255.255.240
B. 255.255.255.254
C. 255.255.255.252
D. 255.255.255.0
E. 255.255.255.248
Answer: C
Explanation
The subnet mask of 255.255.255.252 gives only 2 usable host addresses because it has only 2 bit 0s (22 – 2 = 2) so it is the most efficient subnet mask for point-to-point serial links (and you should remember it).
Question 7
Refer to the exhibit. HostA cannot ping HostB. Assuming routing is properly configured, what could be the cause of this problem?
A. HostA is not on the same subnet as its default gateway.
B. The address of SwitchA is a subnet address.
C. The Fa0/0 interface on RouterA is on a subnet that can’t be used.
D. The serial interfaces of the routers are not on the same subnet.
E. The Fa0/0 interface on RouterB is using a broadcast address.
Answer: D
Explanation
Now let’s find out the range of the networks on serial link:
For the network 192.168.1.62/27:
Increment: 32
Network address: 192.168.1.32
Broadcast address: 192.168.1.63
For the network 192.168.1.65/27:
Increment: 32
Network address: 192.168.1.64
Broadcast address: 192.168.1.95
-> These two IP addresses don’t belong to the same network and they can’t see each other -> D is the correct answer.
Question 8
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 9
If an Ethernet port on a router was assigned an IP address of 172.1.1.1/20, what is the maximum number of hosts allowed on this subnet?
A. 4094
B. 1024
C. 8190
D. 2046
E. 4096
Answer: A
Explanation
In the prefix /20 we have 12 bit 0s so the number of hosts-per-subnet is 212 – 2 = 4094.
Question 10
A network administrator receives an error message while trying to configure the Ethernet interface of a router with IP address 10.24.24.24/29. Which statement explains the reason for it?
A. The address is a broadcast address
B. The Ethernet interface is faulty
C. VLSM-capable routing protocols must be enable first on the router.
D. This address is a network address.
Answer: D
@xallax
which command helps a network administrator to manage memory by displaying flash memory and NVRAM utilization
a.show secure
b.show file systems
c.show version
d.show flash
plz replay
@islam
question 4 from:
http://www.9tut.com/ccna-cisco-ios-questions-3
also read here:
http://www.cisco.com/en/US/docs/ios/12_3/configfun/command/reference/cfr_1g08.html#wp1042676
cheers
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!
Hello 9tut!
Thank you for the posts, they have been really helpful in my studies. I have taken the 640-802 and failed once already, I blame the EIGRP simulation because I haven’t the slightest clue why it wouldn’t work and I wasted 30 min trying to make it work so… thats where I really messed up my chances at passing, I didn’t finish 12 questions on the test due to time.
Would someone send me the latest dump please? I could really use it for studying, my next test is in 3 days. Also if someone could give me a straight answer on the “native vlan 999 question” that seems popular lately I would appreciate it because it is confusing the heck out of me. -Burnout771@msn.com
-thanks to all who contribute!
Hi everyone,
Thanks heaps for all the hard work, I’m wondering if you can guys help me by sending the latest dump so I can get an idea for the exam. alaradi.h@gmail.com
Ok regarding Q8 (113 point to point links) i came accross this same question somewhere else and they concluded the correct answer was B. They started from 32 bits – 7 bits = /25. So the question is should we count back from 32bits or 30bits??? Please share your thoughts??
I agree with you
32-7= /25 is correct.
Sorry
Q8
we need 113 subnets with 2 address ip Usable
32-7= /25
i don’t ubderstand why you make 30-7= /23
Point to point Links – -> /30
so 30-7 – -> /23
These are the host requirement for this network 130.120.0.0:
120/25
76/25
200/24
224/24
700/22
1414/21
2500/20
3220/20
10300/20
10300/18
14/28
30000/17
14000/18
What ip address so i use?
segun_adegunsoye@yahoo.com
These are the host requirement for this network 130.120.0.0:
120/25
76/25
200/24
224/24
700/22
1414/21
2500/20
3220/20
10300/20
10300/18
14/28
30000/17
14000/18
What ip address so i use?
@segun
start off by putting the bigger subnets at the beginning of the network (start with 130.120.0.0/17) then continue with them in descending order.
cheers
Q4 was there in today’s exam.
Q8 (hopefully) clear explanation:
This stumped the hell out of me, but im gonna try have a go to make it as a clear as possible.
warning, you will need basic subnetting for this….
1st: the q is hard because they dont tell you how many hosts they want, and second, VLSM makes this hard to judge since they cant be classed. but have a look at my breakdown (the way i see it) and see if it makes sense.
2nd: clear your mind of all the jargon and crap in the way. no complicated math, binary, etc
**key point to remember, you need 113 NETWORKS, wasting least number of NETWORKS**
A. 10.10.0.0/18 <– this gives you 4 networks, with about 16382 addresses (0, 64, 128, etc)
B. 10.10.0.0/25 <– this gives you 2 networks, with 126 addresses (0, 128)
C. 10.10.0.0/24 <– this gives you 1 network, with 254 addresses (0)
D. 10.10.0.0/23 <– this is the answer, so let me break it down:
/23 = 254, then, 256 – 254 = 2…which means, 256/2= 128 networks,
(or links, exceeding 113, wasting only 15), (dont worry about how many
addresses this gives you)
E. 10.10.0.0/16 <– this gives you 1 network, with about 65023 addresses (0)
and there you have it. links = networks. well this is how it made sense to me. sorry some of the math might be off for the bigger host numbers. just remember amount of hosts dont matter.
hi guys
pls can some one help with question 4
y is is B. 192.168.1.64/26 and not D. 192.168.1.56/26
@ abbey
Looking at the exhibit you can determine that the network address used is 192.168.1.0 and its being subnetted using VLSM for a better addressing scheme.
Router1 is using mask /27 so the increment is 32. So the Router1 is using the following range of ip addresses:
0 – 31
Router2 is using mask /28 so the increment is 16. So the Router2 is using the following range of ip addresses:
32 – 47
Router3 is using mask /28 so the increment is 16. So the Router3 is using the following range of ip addresses:
48 – 63
Router4 needs to encompass 60 host so the only mask that will allow us to do that WITH THE LEAST WASTE of ip addresses is a mask /26 as you already know.
So far you can see that the ip addresses starting from from 192.168.1.0 to 192.168.1.63 are all in use already and we cannot use them.
If you were to use the ip address 192.168.1.56/26 with that mask, this mask allows 4 networks with 64 ip addresses each. This would be the network ranges that mask would encompass:
0 – 63
64 – 127
128 – 191
192 – 255
As you can see the address 192.168.1.56/26 is part of the 0 – 63 range and those ip addresses are already in use by the Routers 1 2 & 3 and therefore we cannot use them anymore since it will overlap.
We need to use the range 64 – 127 that starts where we left off in router 3
Option B. 192.168.1.64/26 is correct
Bcoz 192.168.1.64/26 is a subnet address while the other one is not…In the question they have asked for the subnet address…I hope you got it now…!!!!
question 10 I think the answer is c
hi guys i have seen know question 10 10.24.24.24/29
answer D is corrent
Can Anyone give the answer for the following question?
Q.If HostA pings to S0/0 on Router3,what will be the TTL value for that ping before it enter Router3?
Diagram is like:
HostA —–> connected to Router1[s0/0]——-[s0/1]connected to couter2[s0/0]——-[s0/0]Connected to Router3 —– HostB
Answer:253
Can anyone explain how??
can anyone send me the latest dumps to E-mail:sriram242@gmail.com, i have exam on 30/03/2102. thanks to 9tut is wonderful material for cisco exams
Q1 in exam. Great collection
@Samreen
When a packet starts travelling it has 255 TTL (Time to Leave ) value
Each time it passes a router TTL is decreased by one
So it starts with 255, Router1 makes it 254 and Router2 makes it 253
hi 9tut and xalax…
i would like to ask for your recommended site for subnetting practice.
Thanks.
Hello guys….I need latest ccna dumps…can anyone help me plssss……If someone is having the same then mail me at…..sandeepchile88@yahoo.com
Hello guys….I need latest ccna dumps…can anyone help me plssss……If someone is having the same then mail me at sandeepchile88@yahoo.com
Q-3
Which address range efficiently summarizes the routing table of the addresses for router main?
A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21
Answer: C
Explanation
To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks C is the correct answer.
MY QUESTION HOW TO IDENTIFY FOR subnetted with a subnet mask of /24
PLEASE EXPLAIN ME
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)
Can someone please explain this answer to me only one I missed thank you.
@xallax- Buddy your illustration for the 113 p2p links ques was superb..Owe a lot of respect 2 u…Thnks again 4 helping us with ur simple & easy methods
Hi EVERYONE – I WOULD LOVE TO GET THE LATEST DUMP FROM ANYONE WHO CAN GIVE IT TO ME> Please!!! I am taking the exam on May 15th and really would appreciate it.
Thank you!!!!
erica3025@gmail.com
Erica
I am confused with Q.1and3.
@tosin
the first question is a bit ambiguous, but the subnet mask should help you better choose the answer as 9tut explained. a mask of /27 gives you 30 assignable IP addresses per subnet
regarding 3rd question… we are assuming the subnet mask used for the LANs is /24.
option C is the best pick, but it is not the best way to summarize it as you are also including the .0.0/24 and .14.0~.15.255
still, this is the best of the available options because
A. 172.16.0.0/18
summarizes from .0.0 to .63.255. this is way bigger then the .1.0~.13.255 range that we’re given to summarize
B. 172.16.0.0/16
from .0.0 to .255.255. this is an example of classful auto summarization and it exceeds the given range by a way too much.
C. 172.16.0.0/20
this would also include the above mentioned subnets (.0.0/24 and .14.0/23). it is the best pick as the others are either way too big or not big enough (option D)
D. 172.16.0.0/21
this is summarizing from .0.0 to .7.255. it isn’t big enough to include all the advertised networks that are behind Main Router.
@xallax tanks a lot
thanks 9tut!=D
Please explain to me how to solve this type of questions step-by-step. I feel very confident with subnetting questions but this type of questions for some reason I just cant understand them and know where to start.
The company needs a minimum of 300 sub-networks and a maximun of 50 host addresses per subnet.Working with only one Class B address, which of the following subnet masks will support an appropiate 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
Never mind, I get it now :-)
Q7.
The Answer is C because fa0/0 cant assign that IP address as it overlaps with the se0/0 IP address
@Dimuth
If you say that the cause of the problem is that Fa0/0 has assigned a subnet that CANT be used because it overlaps with se0/0 ip address, then the solution according to you would be to change the subnet of Fa0/0 to a different one that doesn’t overlap with se0/0 ip address.
BUT imagine that you change the subnet of fa0/0 to a different one that doesn’t overlap with se0/0 ip address. The problem will persist because the ip address of the serial interfaces of the routers still are on different subnets:
RouterA se0/0 belongs to —————> 192.168.1.32 to 192.168.1.63 subnet
RouterB se0/0 belongs to —————> 192.168.1.64 to 192.168.1.95 subnet
Answer C just says you cant use subnet 192.168.1.32 on Fa0/0. But that’s not true because there is no restriction nor reason why a network technician would not be able to use a subnet on an interface , with the exception in a scenario where the ip subnet-zero is disabled. And of course following the basic rules of ip addressing.
Option C is incorrect.
The problem was that the IT guy configured the serial interfaces in different subnets. Yes he used an overlapping address, but even if he would’ve used a non-overlapping address, the issue will continue if they are in different subnets.
Option D is correct!!! I hope you find this helpful :-)
Danilin— 50 , the next number is 64 and 300 the next binary number is 512
where 512 is 2 to the ninth and 64 is 2 to the sixth
@geedub
Thanks ;-)
@9tut
Thanks a lot.
Hi all, I am taking CCNA 640-802 exam first time on 30/05/2012. Could anyone please send me latest dumps which are valid for UK? My e-mail address is puneet_gill84@yahoo.co.uk. Many thanks.
How can i calculate or determine the increment of a particular ip address???somone hlp me pls.
@xallax
logic:
each subnet occupies N IPs. i need 113 subnets with N IPs each.
113 * N < 2^P
for the last example, 1022 hosts, my logic was:
i need 113 subnets with 1024 IPs (1022 + 2)
113*1024 >>> should it be /15, 255.254.0.0 ( 113 subnets with 1024 IP’s ) ???
solution: 1024 = /22 = (22 bits)
113 < 128 (128 = 2^7) = (7 bits)
22 bits – 7 bits = 15
/15 = 255.254.0.0
correct me if im wrong :-)
@xallax
copy paste is not complete but that’s what i mean, in your previous post :-)
I PASSED CCNA EXAM TODAY THANKS TO ALL MIGHTY ALLAH
960/1000
@pelikan
113 subnets of 1024 IPs (we count the subnet and broadcast IPs too, very well pointed out)
here’s my logic…
113 * 1024 is… some number… how many bits does that need when subnetting?
1024 is 2^10 and
113 is greater than 64, less than 128.
we use the larger one, 128. 128 is 2^7
so… 2^10 * 2^7 = 2^17 for hosts, that leaves 2^32 : 2^17 = 2^15, 15 bits for the subnet mask.
the answer is /15 (255.254.0.0)
i took the extra step and made the math there, you took the very short path. very well sir :)
@xallax dude u are a genius.have fully grasped subneting now so was wondering wats the next topic to do after subneting .am self studying using my fone and subneting was my second topic after the OSI.
@dougenius
try converting the cbt nuggets vids to the correct size or format for your phone and watch them as you go. they can make the difference
To all and xallax
can you please help me with this..
What is the summarized for these network
172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
my answer id 172.1.0.0/21 but in the dump it’s 172.1.0.0/22 please help me with this..thank you so much for your help..