ICND1 – Subnetting
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Premium Member: You can test your knowledge with these questions first via this link.
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Note: If you are not sure about Subnetting, please read my Subnetting Made Easy tutorial.
Question 1
[am4show have=’p2;’]What is the subnet address for the IP address 172.19.20.23/28?
A. 172.19.20.0
B. 172.19.20.15
C. 172.19.20.16
D. 172.19.20.20
E. 172.19.20.32
Answer: C[/am4show]
Explanation
From the /28 we can find all information we need:
Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)
In fact we don’t need to find out the broadcast address because the question only asks about subnet address (network address).
Question 2
[am4show have=’p2;’]What is the network address for the host with IP address 192.168.23.61/28?
A. 192.168.23.0
B. 192.168.23.32
C. 192.168.23.48
D. 192.168.23.56
E. 192.168.23.60
Answer: C[/am4show]
Explanation
From the /28 we can find all information we need:
Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 192.168.23.48 (because 48 = 16 * 3 and 48 < 61)
Question 3
[am4show have=’p2;’]Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
A. 192.168.1.8/29
B. 192.168.1.32/27
C. 192.168.1.40/29
D. 192.168.1.16/28
E. 192.168.1.48/29
Answer: C[/am4show]
Explanation
From the subnet mask of 255.255.255.248 we learn:
Increment: 8 (248 = 11111111.11111111.11111111.11111000)
Network address: 192.168.1.40 (because 40 = 8 * 5 and 40 < 42)
Question 4
[am4show have=’p2;’]Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)
A. 10.1.168.0
B. 10.1.176.1
C. 10.1.174.255
D. 10.1.160.255
E. 10.1.160.0
F. 10.1.175.255
Answer: A C D[/am4show]
Explanation
From the /20 we can find all information we need:
Increment: 16 (/20 = 11111111.11111111.11110000.00000000). This is applied for the 3rd octet.
Network address: 10.1.160.0 (because 160 = 16 * 10 and 160 = 160 -> the IP address above is also the network address.
Broadcast address: 10.1.175.255 (because 175 = 160 + 16 – 1)
Therefore only 10.1.168.0, 10.1.174.255 and 10.1.160.255 are in this range. Please notice 10.1.174.255 is not a broadcast address and can be assigned to host.
Question 5
[am4show have=’p2;’]Which one of the following IP addresses is the last valid host in the subnet using mask 255.255.255.224?
A. 192.168.2.63
B. 192.168.2.62
C. 192.168.2.61
D. 192.168.2.60
E. 192.168.2.32
Answer: B[/am4show]
Explanation
Increment: 32 (224 = 11111111.11111111.11111111.11100000)
Network address: x.x.x.(0;32;64;96;128;160;192;224)
Broadcast address: x.x.x.(31;63;95;127;159;191;223)
-> Last valid host (reduced broadcast addresses by 1): x.x.x.(30;62;94;126;158;190;222) -> Only B is correct.
Question 6
[am4show have=’p2;’]An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two)
A. 192.168.4.61
B. 192.168.4.63
C. 192.168.4.67
D. 192.168.4.125
E. 192.168.4.128
F. 192.168.4.132
Answer: C D[/am4show]
Explanation
Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has been subnetted with a /26 mask so we have 2(26-24) = 22 = 4 sub-networks:
1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)
In all the answers above, only answer C and D are in the same subnet.
Therefore only IPs in this range can be assigned to hosts.
Question 7
[am4show have=’p2;’]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[/am4show]
Explanation
With network 192.168.20.24/29 we have:
Increment: 8 (/29 = 255.255.255.248 = 11111000 for the last octet)
Network address: 192.168.20.24 (because 24 = 8 * 3)
Broadcast address: 192.168.20.31 (because 31 = 24 + 8 – 1)
Therefore the first usable IP address is 192.168.20.25 (assigned to the router) and the last usable IP address is 192.168.20.30 (assigned to the sales server). The IP address of the router is also the default gateway of the sales server.
Question 8
[am4show have=’p2;’]Given a Class C IP address subnetted with a /30 subnet mask, how many valid host IP addresses are available on each of the subnets?
A. 1
B. 2
C. 4
D. 8
E. 252
F. 254
Answer: B[/am4show]
Explanation
The number of valid host IP addresses depends on the number of bits 0 left in the subnet mask. With a /30 subnet mask, only two bits 0 left (/30 = 11111111.11111111.11111111.11111100) so the number of valid host IP addresses is 22 – 2 = 2. Also please notice that the /30 subnet mask is a popular subnet mask used in the connection between two routers because we only need two IP addresses. The /30 subnet mask help save IP addresses for other connections. An example of the use of /30 subnet mask is shown below:
Question 9
[am4show have=’p2;’]Which two statements describe the IP address 10.16.3.65/23? (Choose two)
A. The subnet address is 10.16.3.0 255.255.254.0.
B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
E. The network is not subnetted.
Answer: B D[/am4show]
Explanation
Increment: 2 (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0)
Network address: 10.16.2.0 (because 2 = 2 * 1 and 2 < 3)
Broadcast address: 10.16.3.255 (because 2 + 2 – 1 = 3 for the 3rd octet)
-> The lowest (first assignable) host address is 10.16.2.1 and the broadcast address of the subnet is 10.16.3.255 255.255.254.0
Question 10
[am4show have=’p2;’]What is the subnet address of 172.16.159.159/22?
A. 172.16.0.0
B. 172.16.128.0
C. 172.16.156.0
D. 172.16.159.0
E. 172.16.159.128
F. 172.16.192.0
Answer: C[/am4show]
Explanation
Increment: 4 (/22 = 11111111.11111111.11111100.00000000)
Network address: 172.16.156.0 (156 is multiple of 4 and 156 < 159)
Thoz dumps are for free the last time I checked so I don’t know why you had to buy
In Q7 A is also correct – 8 x 1 + 1 etc first host 14 last host 15 broadcast?
In Q5 there actually 2 good reponses B and C : 61 is in the subnet 192.168.2.32/27 and 62 too
In Q5 I didn’t see it but D is also in 192.168.2.32/27 : 60 is a valid host address!
sory I didn’t see the word LAST in the question… the onlyy good answer is B …
in question #10 why is 156.0 ….?? I do not understand
Regarding last comment: the prefix is /22 so the mask in binary is 11111111.11111111.11111100.00000000
Wanttopass, no, the A is not correct in Q7 because we have to choose the first usable host address for router and the last usable host address for server in subnet 192.168.20.24/29.
@joel, answer is right!
In question 7, A and C are correct:
Answer A
Network: 192.168.20.8
Router first usable. 192.168.20.9
Subnet Mask: 255.255.255.248
Server last usable: 192.168.20.14
Broadcast: 192.168.20.15
Next network: 192.168.20.16
No. Just look at the Question 7 carefully:
“For network 192.168.20.24/29, the router is assigned the first usable host address”
/29 in DDN is 255.255.255.248 = 8 address per subnet.
192.168.20.24/29:
Network address: 192.168.20.24
First usable: 192.168.20.25, this is gateway
last: 192.168.20.30, this is server
Broadcast: 192.168.20.31
Did you got it?
please help me i want new question in ccna 120-200
thanks for all
In Q7 explanation….how to know there are 8 subnets from 255.255.255.248 ???
sorry, i mean how to know there are 8 addresses per subnet ….thanks
256-248 = 8
subtract the value of the last incomplete byte from 256 and that’s the short and fast answer
hiii,
In Q 10, how is it 156??
192 & 128 also are multiple of 4 as well.!!!
Please explain.
thanks in advance!!!
If you need help learning to subnet, etc. Here is a video that I used to refresh and/or learn from:
http://www.youtube.com/watch?v=rs39FWDhzDs
Q10: Does not make sense. Please explain in greater detail. Thanks!
Anyone,
I plan on taking exam next week. What other sims/labs did you get besides the OSPF 6 router?
Thanks in Advance!
The answers for Question 4 are: C. Just that–only C. Read the question carefully:
Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)
The key is “for hosts”. They’re asking for host addresses. All other answers are network or broadcast addresses. You can’t choose three, because only one is an address that is valid for a host.
OK, I take it all back. The light went on: the correct answers are A, C, D. Explanation is good. 2 to the twelfth power (32-20=12), is 16x(2 to the eighth), which means the range is 160.0 through 175.255. Valid host addresses are 10.1.160.1 through 10.1.175.254.
Q6 ?
192.168.4.0 / 26
192.168.4.1
192.168.4.62
192.168.4.63
Why C and D ?
For the one who said A and C are correct for question 7 your very wrong,mind you that you have given the network id of 192.168.20.24/29 so the correct answer was C,the question is so trick so you have to be very carefully ,thanks for the questions they have been important to me
@brahmanand although 192 and 128 are multiple of 4 but are not the networks id for the given host(172.16.159.159/22) so obviously the answer will be 172.16.156.0/22,i hope it has been informative to you friend
question 6
the subnet ID 0 64 128
the broadcast network 63 127 255
so only valid subnets are 65-126
therefore C. 192.168.4.67
D. 192.168.4.125
are these questions the exact questions on the test? Meaning are the numbers the same here as they’re on the test?
Even if these are the exact questions on the test, it makes no difference. You need to truly understand subnetting for this test. If they change one octet or even a few numbers, you will have to truly subnet.
Have a question regarding question 10. I understand that the subnet (Network ID) is 172.16.156.0 and that 172.16.159.159 is part of the subnet with the next subnet being 172.16.160.0 and broadcast 172.16.159.255. So that would give 2 usable addresses 172.16.157.255 and 172.16.158.255. Is my thinking right or am I slightly off course. Any help please as to whether I am on the right path. I do understand that /22 gives 255.255.252.0 and an increment of 4 (-) 2. One address for the Network ID and One for the Broadcast which would give 2 usable addresses. Does the last octet appear as .255 as we have not subnetted into the fourth octet. I think I am on the right path but any help would be appreciated. Thanks Chris
@Chris
Simple rule:If the address has 255 at last octet,it doesnt mean that it is broadcast IP.
Thanks Konan. So in the above question 10 does that mean we use 255 for the last octet? We subnetted in the 3rd octet. What would be put in the fourth octet? Thanks for any help. Chris
I just remember the rule, that if all 0’s then it is the Network ID and if all 1’s then it is the Broadcast.
All the answers are as they say they are. Best part about it for me is that I figured them all out in my head. It’s really very simple if you find the right study material. I have only been studying for about a week now. But I do know some things about computers and network. Pertaining to Q7. The subnet is 24 next one starts at 32. 31 is the broadcast so 30 is the last available host in that subnet you can use. Since 24 is the subnet the 1st host is 25. See how easy that was?
Hello,
Would somebody explain question 4.
Why 10.1.176.1 is couldn’t be a valid host address in 10.1.160/20 network ?In my opinion 10.1.176.1 is a part from 10.1.176.0 – 10.1.191.255 network {10.1.176.1 – 10.1.191.254 valid host}
@ico,
Q4
Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)
/20 = 16 block size
160 = multiple of 16
so you would think the host range would be 10.1.160.1 to 10.1.175.255 BUT you need to take away the broadcast address.
so the real range of hosts is 10.1.160.1 to 10.1.175.254..
and the next network (block size 16!) is 10.1.176.0
thus the answers being A, C, D
A. 10.1.168.0 (valid host)
B. 10.1.176.1 (in the next subnet!)
C. 10.1.174.255 (valid host)
D. 10.1.160.255 (valid host)
E. 10.1.160.0 (network address)
F. 10.1.175.255 (broadcast address)
Thank you Ron. I got it.
There was my mistake.
I just want to ask if in the exam they are going to have questions that are large numbers to subnet like 186.218.196.42/20? The increment is 16 here
Answer to question 7 is C. Read the question properly, it is asking For network 192.168.20.24/29….No the subnets that are created using 0,8,16,24,32…..ETC.
For network 192.168.20.24 Subnet the first usable is .25 and last is .30 (.24 is network and .31 is broadcast).
Hope this help to understand.
Harsh
I do not understand question7
Passed my 100-101 exam today and question 1,2,3,4,5 and 6 were there.. all the best
on Q6 , the 4th subnet range should be
4th subnet: 192.168.4.192 (to 192.168.4.255)????
the explanation might confuse people please fix this! it is a /26 so the increment is 64
Currently you have posted
1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)
Got 986/1000 Marks, {5/16/2014} 50 Questions, Ospf Sim, Security simlet, and Router and switch Simlet Was also there…..All questions From 9tut. and Examcollections…..Now Heading For Icnd2 And Blogging…..
For Any Help.
Manohar Tn
Website: http://www.techlinko.com
G+ : https://plus.google.com/u/0/+ManoharTN9/
Question Six .125 does not fall under .127 there is a mistake there.
Nevermind it does fall under. so it’s OK
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What is the network address for the host with IP address 192.168.23.61/28?
A. 192.168.23.0
B. 192.168.23.32
C. 192.168.23.48
D. 192.168.23.56
E. 192.168.23.60
I think answer B. If increment is 16….32…64..128…. Why 48 ?…………….. Please explain
can any 1 explain question 10…..
^typo error 192.168.23.17 should be 192.168.23.15
guys check this link… Very good video on ip addressing and sub netting
https://www.youtube.com/watch?v=FrdaCx3R9CM
Please may you explain in detail question no 9?
http://i.imgur.com/eylHQ1P.jpg
/x = CIDR Mask
Blue # = DDN mask in interesting octet.
Green # = Magic Number
Remember this chart for exam if you have trouble working the mental math out.
Anyone seen any VLSM questions on the exam?
q4? 10.1.160.0 is not valid host. its a network address
Question 2: What is the network address of 192.168.23.61/28?
A bit misleading… The network address is 192.168.23.0
As for the subnet address, then 192.168.23.48
Hi Fox,
I need a help on the way i could respond on this type of subnetting question in less than 2 minutes: What is the subnet address of 172.16.159.159/22?
I know that this is a class B address and the default mask is 255.255.0.0
Also, I know that /22 is 255.255.252.0 subnet mask and the magic number is 4
what can be the fastest way to found out that 159 on the third and fourth octect is a broadcast address without going by counting the increment 4(0,4,8,12,16….156,160…). Thanks for your help.
Thanks you very much Lee for the clue in subnetting questions.
Good and fast way to subnet.
Best regards,
i dont understand question number 8
The Answer of Q10 is Wrong.
Question 10
What is the subnet address of 172.16.159.159/22?
A. 172.16.0.0
B. 172.16.128.0
C. 172.16.156.0
D. 172.16.159.0
E. 172.16.159.128
F. 172.16.192.0
Answer: C – WRONG!!!!
Explanation:
IP: 172.16.159.159/22 or 255.255.248.0 – 256-248 = 8 in 8 on the third octet.
156 is not a multiple of 8.
Subnet:
172.16.152.0 – 172.16.159.255
IP 172.16.159.159 belongs to 172.16.152.0/22 subnet. You can check on online ip subnet calculator.
Becareful!
Sorry my comment above. Is /22 not /21 Q10 – Answer C is right.
Sorry again.
Q 10 has 172.16.159.159/22
/22 means 255.255.252.0
as there are six bits on in second octet so 2exp6=64 subnets and 11111100=252
now total of 64 subnets ….256-252=4 so 0,4,8,12 so on so multiple of 4
and 159.159 subnet should be 172.16.156.0
hope it helps
my test is on 22nd of dec, any body wants to group study for subnetting ?
Sure. How do we do it. I am doing the exam on Thursday tho
@deriq
mine is on monday 29th, ccent brainbump from anyone? appreciations.
pls email braindump to abrahamshen@rocketmail.com Thanks-a-million!!
Q10 is still confusing .. im using the method jeremy taught on cbtnuggets anyone doing it that way? Please help. Thanks.
test
MONDAY HERE I COME ICND1
hi
can i have a CCENT material and which one is eaiser CCENT or CCNA
thanks
Tara are you trolling or just retarded??
LOL Tara and WTF
Q4 isnt the first valid ip for .160 network .160.001? So 160.255 a vaild ip? Im might be missing something.
any ccent practice test questions
For those asking about Q10:
What is the subnet address of 172.16.159.159/22?
A. 172.16.0.0
B. 172.16.128.0
C. 172.16.156.0
D. 172.16.159.0
E. 172.16.159.128
F. 172.16.192.0
When looking at the octets, know where the /masks are
(/8) (/16) (/24) (/32)
(172) . (16) . (159) . (159)
So when we see a /24, we are dealing with the fourth (last) octet. HOWEVER, in this question it asks for /22 so we are in between /16 and /24, meaning we are dealing wit the third octet. Most of you know your fourth octet subnet sizes but get confused when it goes to octets prior to that. Think of a /22 as a /30, which we know has a subnet size of 4.
So if we know that the subnet size is four, that’s a little easier to deal with. Now we can (for now) ignore the fourth octet and just look at the increments of four to see our network addresses:
172.16.0.x
172.16.4.x
172.16.8.x
…all the way down to
172.16.156.x
172.16.160.x
each of these subnets have four numbers in the third octet but they also have 255 numbers in the fourth octet as they count up
Subnet #1:
172.16.0.1
172.16.0.2 etc
Subnet #2:
172.16.4.1
172.16.4.2 etc
We know that our address given is 172.16.159.159, so if we ignore (temporarily) the fourth octet, we can see that 172.16.159.x falls in between 172.16.156.x and 172.16.160.x
172.16.156.0 <— (The subnet address)
172.16.156.1
172.16.156.2
…all the way down to
172.16.159.158
172.16.159.159 <— (Here it is in the same subnet!)
172.16.159.160
…all the way down to
172.16.159.255 <—(The broadcast address of the subnet)
Hope this makes a little more sense, as dealing with octets other than /24 and above can get confusing.
Q4…baa and slappypappyj…..remember every subnet in this example has 4094 hosts….that’s the trick and what is throwing both of you…it’s easier if you write this one out in binary
q4,8 in icnd1 exam
Can someone please tell me how many TOTAL subnetting questions we can expect in exam? 5? 10?
hi!
Could you please clarify on Q3 – why the answer is C?
@Nino
Q3 Because 255.255.255.248 248 Mask is a /29
/29 = 8 hosts ( 6 valid) so the IP address is a .42 so we know its part of the .40 subnet
answer is C:. 192.168.1.40/29
@Nino
Q3 Because 255.255.255.248 = 248 Mask is a /29
/29 = 8 hosts ( 6 valid) so the IP address is a .42 so we know its part of the .40 subnet
answer is C:. 192.168.1.40/29
i will be taking my exams and these make me feel confident
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Q6 (Which two addresses can be assigned to hosts within the SAME subnet?), aren’t C & D in two separate subnets, I believe the question refers only to subnet 192.168.4.0/26 which involves a range of addresses of 4.1 to 4.62, only A and B fall within this subnet but B is a broadcast so not usable for hots, the rest are all in different subnets, unless the question refers to 192.168.4.0/24 then all of them are part of that network.
@all: We had to move all the questions and answers out of 9tut. We can only keep the explanation. You can download the questions and answers at: https://mega.co.nz/#!oIdESYbD!yyu33vygrfKPy4rcmcbV6qW2fxINNoTokuDM3CjA_og
Hi everyone! How come i do not see questions for ICND2.. My membership might have expired but i can’t get to my profile to re new it..
@Liliya: You can access your profile at https://www.9tut.net/member/profile
hi all.
I took ICND1 23-07-2015 and passed 907/1000………….thanks 9tut. I only study this site and got 907/1000 so, if you study extra sites you will be better than me. yeh there is no any configuration only practice “show COMMANDS” and practice all the SIMs.
Hi,
Just passed the exam last week and there was a lot of subnetting, mostly Class A and B. Please watch this video because it did help me;
https://www.youtube.com/watch?v=rFra2TtBrk4
how can i get question of dis ans???? plzz help
hi guys can help me with this ,Network address: 192.168.1.40 (because 40 = 8 * 5 and 40 < 42) why and how to get the '8 * 5' ????? why *5 ?? Q3
Question 6
Explanation !!! can’t understand at all pls help out guys
guys for question 4 I am getting broadcast as 10.0.191.0/20
Can someone please help where I am going wrong? I got 10111111 as the 3rd octet
@charles. I too was confused until I took a triple look. There are four possible subnets that can be created.
192.168.4.0/26
192.168.4.64/26
192.168.4.128/26
192.168.4.192/26
What the question is basically asking is, which of these four subnetworks holds two of these hosts addresses. The answer is C & D because the host range of the 192.168.4.64/26 network is 192.168.4.65-.126/26
Hope this helps you or anyone else that is confused.
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I really am confused here:
“An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two)”
It states 192.168.4.0 ! (so not: 192.168.1.64). So in my honest opinion, the question is noted wrong…
Am I correct?
ow, never mind, already see the comment of Bossaver now 🙂
285q dumps are still valid! Thanks tut and everyone sharing here
In question 6 the first subnet is 192.168.4.0-192.168.4.63,
Why isn’t Answer A 192.168.4.61 be a host in that subnet?
Why i don’t see any questions? I can see only the answers.
Thanks @Armando 285 still good