Subnetting Questions

9tut, can you please explain the answer for question 1 , for example (A)interface fa0/0 has an ip address that overlaps with network 10.1.3.0/30.

Thank you.

The IP address of fa0/3 is 10.1.3.1 which belong to network 10.1.3.0/30 -> interface fa0/3 has its IP address overlaps with that network.

hi..
to question 3.
why the answers is 192.168.1.190 ..
Did not 192.168.1.64?
how is it calculated?
thx…

the answer to question 3 is done by 32. So to answer the question you are borrowing 3 bits to bet 224. Add 128+64+32=224.

Q.3 /27 or 224 = 32 addresses
networks 0,32,64,92 etc

therefore 64 is a network address and cant be used.. the only address usable is 190. the rest are either network or broadcast

why not .192 in the case of question 3…..can some1 explain please…thx.

.192 is the network address and it is not assignable to an interface.

Q1:
A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30.

nothing there used that subnet aside from fa0/1

please clarify Q1, i dont get it why A is one of the answers

The IP address of interface Fa0/3 is 10.1.3.1 and it belongs (overlaps) to network 10.1.3.0/30

9tut…other than fa0/3 what other interface is using 10.1.3.0/30 subnet? I dont see that n d diagram.

Only Fa0/3 belongs to 10.1.3.0/30 subnet.

A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30 makes no sense. There is no mention of 10.1.3.0/30 anywhere in the diagram. I could understand that IP address for Fa0/3(10.1.3.1) does not belong in the subnet 10.1.3.128/25 but in subnet 10.1.3.0/25.

The question just wants to test if the candidate understand the IP address 10.1.3.1 belongs to 10.1.3.0/30 or not. Although I agree with you that this question is a bit unclear but we have to answer it anyway.

A better explanation of why Question 1 answer A is a problem. – the diagram shows that the Router Fa0/3 and the switch should share a common LAN Subnet 10.1.3.128/25. Denoted by the circle.

Router Fa0/3 IP Address 10.1.3.1/25 = Subnet 10.1.3.0 (IP address range 10.1.3.1 to 10.1.3.126) broadcast 10.1.3.127.

The Common LAN subnet IP 10.1.3.128/25 = subnet 10.1.3.128 (IP address range 10.1.3.129 to 10.1.3.254) broadcast 10.1.3.255

They (Rtr & Switch) are not sharing the same subnet and cannot communicate based on the information presented.

Can Question 1 B and D be explained as ,
for B,
Subnet Address – 10.1.1.0
Broadcast Address – 10.1.1.3
Next subnet – 10.1.1.4

So since the IP address assigned is meant for the next subnet

for D,

Class A – 17 Subnets bits borrowed – 2^ 17 subnets

7 bits for host, so 2^7 -2 = 126 host bits

mogambo, you have too look closely and see between the routers (RouterA too RouterB the serial link).

Look closely on the ip’s and see that they are on different subnets.

/27 = 2^5 = 32 increment

subnets:
0-31
32-63 RouterA
64-98 RouterB

Question 4.
I fail to see why answer D is correct.

It states routing is properly configured. I assume either static or dynamic, so the route should be learnt via the protocol. Am I missing something else. ???

Any help appreciated.

Cisco_kid:

When two routers are connected via serial, they have to be on the same subnet.

The IP of the left router is 192.168.1.62 while the other is 192.168.1.65 ; both share a mask of /27 = 255.255.255.224.

There are two ways to solve this problem, the long way is to use binary, and calculate [LeftIP AND 255.255.255.224] which results in network 192.168.1.32. Then do the same for the right IP, which results in network 192.168.1.64. You can see the networks are different here.

The other option is to realize the increment is 32, so your subnets go from 0…31, 32…63, 64…95, etc. Then you can see that .62 and .65 on different subnets.

To clarify why they need to be on the same subnet, it’s because directly connected parts of a network need to be on the same broadcast domain so they can “see” each other. I’m sure someone has a more technical / clear explanation, but that’s the way I think of it.

THIS SITE IS REALLY HELPFULL .EXPECIALLY AFTER QUESTION IT DESCRIBE VERY WELL..QUESTION .I have got 1 question ..192.168.1.5/30…then how can we assign to host this question came in 9sep 2010..tell me how…

For Q1/A :
“Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30″
is CORRECT “per se” ( where overlaps === belongs to ) … but should be replaced with:

“Interface fa0/3 has an IP address that DOES NOT BELONG to network 10.1.3.128/25″
or
“Interface fa0/3 has an IP address that SHOULD FORCE THE NETWORK TO WHICH IT BELONGS TO HAVE AN IP 10.1.3.0/25″ ( instead of 10.1.3.128/25 )

Regarding Q1/B … “Interface fa0/1 has an invalid IP …”
Once again , the IP address IS NOT invalid “per se” … it is just un-suitable with the network ( as I stated above )
A really INVALID adress could be : 10.11.345.68

In my opinion … Q1 should have an additional answer like this :
G) Every exam has its own stupid Q/A(s) that create(s) productive debates

Best regards.
Cosmix

i dont understand ,how no 2,get the answer?

to samuraC

2^12 – 2= 4094
n-network bits
h-hosts bits
nnnnnnnn.nnnnnnnn.nnnnhhhh.hhhhhhhh /20 = 12 host bits

@ssss

would the exam expect us to work that out without a calculator? :O

Can someone explain Q 1 answer F – The IP subnet 10.1.1.0/30 is invalid for a segment with a single server.

Thanks

Q1 answer F is incorrect as they are attempting to trick you into thinking /30 is only used for point-to-point links; however, it is perfectly valid to create a /30 subnet where only one device will occupy the address space.

Remember:
/30 contains only 2 usable addresses for HOSTS.

Edit to last post:
“it is perfectly valid to create a /30 subnet where the server occupies 1 address and the router’s interface occupies the 2nd address space”

Here’s the breakdown (had it been configured correctly from the start):

Previous Subnet: 10.1.1.0

Current Subnet 10.1.1.4
Router Fa0/1: 10.1.1.5
Server: 10.1.1.6
Broadcast: 10.1.1.7

Next Subnet: 10.1.1.8

I think that covers things.

simple but time consuming questions ………………..lol

256-224 = 32

192.168.1.31 = Broadcast address – 192.168.1.0 and 192.168.31 are the subnet and broadcast address for this network.
192.168.1.64 = Subnet address – 32×2 = 64 is a subnet address
192.168.1.127 = Broadcast address – 32×4 = 128 is the subnet address. 128-1 is the broadcast of previous subnet which 192.168.1.96 (32×3) subnet.
192.168.1.190 = RIGHT ANSWER!
192.168.1.192 = Subnet address – 32×6 = 192 which is a subnet address. Broadcast address for this subnet would be 192+32(magic number) = 224, 224-1 = 223 —> IS THE BROADCAST FOR THIS SUBNET.!!!

Hope my explanations are clear enough for understanding

Hello 9 Tut,
can you explain question 1. Answer B (B – Interface fa0/1 has an invalid IP address for the subnet on which it resides.
)
I have done the subnetting: 10.1.1.0/30 and I get a valid host range of: 10.1.1.1 to 10.1.1.6
so 10.1.1.4 i within the valid hostrange.
Please I need help to understand this one. am I doing the subnetting wrong?
I would appreciate the help.

10.1.3.1/25 is valid for THAT subnet. But the subnet is 10.1.3.128/25. So fa0/3 is not on the subnet identified as the subnet for the network…hence it overlaps ANOTHER subnet.

Hello Mac,
sorry, I still dont understand it, are you speaking about 1. Answer B (B – Interface fa0/1 has an invalid IP address for the subnet on which it resides?
or A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30 ??

I am talking about question 1. Answer B.
Thanks

Spainvasio, MAC is refering to A…

Question 1 – Answer B – Interface fa0/1 has an invalid IP address for the subnet on which it resides.

Explanation:

10.1.1.0 /30
255.255.255.252
11111111.11111111.11111111.11111100
increment bit is 4

therefore;
10.1.1.0-10.1.1.3
10.1.1.4-10.1.1.7

.0 and .4 is network address
.3 and .7 is broadcast address

the acceptable IP address for this subnet should be 10.1.1.1 OR 10.1.1.2

hope this is easy for you to understand…

Hello skyyyyy2001,
Thanks so much for your explanation, I now understand it.
I really appreciate the solution you provided.
Can you also explain from Question 1 Answer A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30. ?
I still dont have that one cleared.
Just an update on my studies, I finished watching CBT nuggets already, going thru the bryan advante ebook,also doing practice test and lab from boson, also downloaded cisco packet tracert, very cool tool!
If all goes as planned I will be going for the ICND1 exam at the beginning of June.
Kind Regards.

Hi Spainvasion,

Question 1
Answer A – Interface fa0/3 has an IP address that overlaps with network 10.1.3.0/30

Very misleading for this question, just think “belongs = overlap” for this question as someone has pointed out. In short, fa0/3 IP address is wrong… Why is it wrong? Lets do some calculation;

Subnet mask: 10.1.3.128 /25
/25= 255.255.255.128
binary:1111 1111.1111 1111.1111 1111.1000 0000
increment=128

subnet mask range:
*10.1.3.0-10.1.3.127 (10.1.3.1 will be a valid IP address if the subnet mask is 10.1.3.0/25;because it falls under the first host IP address)

**10.1.3.128-10.1.3.255 (10.1.3.128 is the subnet mask listed on the question;therefore IP address will be valid if its coming from this range, ie: 10.1.3.129-10.1.3.254)

I hope this helps…

Answer for Q3:

Subnet Mask Range for Router 1 is 192.168.1.32 – 192.168.1.63 and Subnet Mask Range for Router 2 is 192.168.1.64 – 192.168.1.95.. Therefore both subnet range dont fall into same subnet.
..

/27 = 32bit it means, incremet is 32.
.
192.168.1.0 – 192.168.1.31
192.168.1.32 – 192.168.1.63 – > Subnetwork Range for Router 1
192.168.1.64 – 192.168.1.95 – > Subnetwork Range for Router 2
.

Hope that helps….

wow… cant believe the trickery on question 1.

the answer A is so ridiculous. the way they worded it is so weird. sure am glad i came here so i dont get the wool pulled over me on the exam

Q1, answer B, was a little mind boggling. Looked at it last night and it didn’t hit me until early this morning. skyyyyy2001, thx for explaining it in detail.

you are welcome my friend

In Q1 it’d be much clearer if they asked whether “Interface Fa0/3 (10.1.3.1) BELONGS to the network 10.1.3.0/30″.

The “OVERLAPS” term usually has the meaning of some unwanted situation, while the “BELONGS” verb suits better this case.

I passed tonight with 975/1000. Questions 3 and 4 were there. Almost the same ones.

Passed with 920 today. NONE of these questions were on it. Does not really matter because the IPs would be different anyway so you still have to do subnetting. If you plan on taking this exam you need to at least be able to do class C subnetting in your head. Most of the subnetting in ICND 1 is with class C addresses so its EASY!

Passed ICND1 today. I had about 3 or 4 of these questions on it. Thanks 9tut. I will donate in a few minutes. I appreciate the site.

Passed ICND1 today with 962…..Didn’t have any of these questions on my exam.

Studied with:
Odom ICND1 Book
Todd Lamle CCNA book
Chris Bryant (thebryantadvantage.com)

Guyz i dont see question on this forum being updated or is it because the exam has not changed

Can any body explain further why the subnet bits for Q2. cant be 16 but rather 12

If an ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A – 1024
B – 2046
C – 4094
D – 4096
E – 8190

look at the subnet mask, it’s /20
the maximum is /32
you subtract 20 from 32 and you get 12 (32-20=12)

maximum number of assignable IPs is: 2^12 – 2 = 2^10 * 2^2 – 2 = 1024 * 4 – 2 = 4096 – 2 = 4094.

Thnaks xallax i was looking for an option to retract my Question….ba thnakx too because u have thaught me a new method to solve this. That was a quick response….

Thnaks xallax i was looking for an option to retract my Question….ba thnakx too because u have thaught me a new method to solve this. That was a quick response….

need help with question 1 answer A. don’t under stand why it overlap/belongs to 10.1.1.0/30. the ip is 10.1.3.1??????

hope you dont mind me posting a link. its a load of subnetting questions. and since you should really know how to subnet (rather than just knowing the answers to a select bunch of questions) i thought they might broaden your subnetting skills. it did for me =-)
http://www.subnettingquestions.com/

is there any other expanition to question 1 no A

i took both icnd1 and 2 today

None of the questions here came up.

passed with 962 and 986.

Thank you all at 9tut.

why do my comments get deleted???????????????

Hi All, has anyone sat the exam recently? was there many questions on IPv6?

Hey guys, nice work here, its been really helpful, n I thank u all at 9tut. Meanwhile, I intend writing my ccna 640-802 exam in a few days, can any one please send me recent dumps? I would really appreciate it, my email add. is – toluene001@yahoo.com.
Thanks.

Can somebody please explain question 2 thanks

@hamid
the IP address is 172.16.112.1/20

from that we find out that the subnet mask is 255.255.240.0 (or /20)
how many bites used up by the network? 20
how many bites left for hosts? 32 – 20 = 12 bites

number of assignable IPs 2^12 – 2 = 4096 – 2 = 4094 (option C)

Hi pls help explain this.

Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0 255.255.254.0?

@gure

172.19.0.0 255.255.254.0
this is 172.19.0.0 /23

class B network
16 bits for full network, 23 – 16 = 7 bits for subnetting
2^7 = 128 subnets (subnet-zero is enabled by default)

32 – 23 = 9 bits for hosts
2^9 – 2 = 512 – 2 = 510 maximum hosts on each subnetwork

can some one explain me how should i start preparation of CCNA

@mustafa
start off by reading todd lammle’s book
http://books.google.ro/books/about/CCNA_Cisco_Certified_Network_Associate_s.html?id=CNQjN-CNPZQC&redir_esc=y

can u explain how 3rd question will done

Subnetting is a must know for the exam. About 60 percent of all questions are relating to subnetting. Choose a method and be good at it. That’s what I do.

Monica , the 3rd question is lemon squeezy . The subnet mask ends in .224 , so, three bits were borrowed . 128+64+32 = 224 , so , therefore the range starts at 32 , and goes up in increments of 32 , so 64 will be a network address , 127 will be a broadcast , as 128 is a new network address , 192 is also a network address , so , .190 is a valid host address , .191 is a broadcast , so is .31 . Understood ?, if not , please say so , as subnetting is my baby .

For subnetting , its practice and practice , until one day , you will find yourself subnetting in your head . For example , for a class c address with mask 255.255.255.240 , means 4 bits were borrowed , and the range starts at 16 and goes up in increments of 16 . So valid addresses for the first range will be .17 – .30 as the number before new subnet is a broadcast and is invalid , so .31 is invalid