Comments on: CCNA – Subnetting Questions 4

By: subnetting

subnetting — Tue, 31 Mar 2015 04:44:11 +0000

from 192.10.20.0/24 come up with four subnets

By: Anonymous

Anonymous — Tue, 31 Mar 2015 04:40:39 +0000

from 192.10.20.0/24 come up with four subnets

By: paboy

paboy — Sun, 29 Dec 2013 13:06:56 +0000

113 point-to-point links. We all know point-to-point links uses a /30. Therefore, 113 point-to-point links is = 113 /30s. We also know in a /30 we have 4 as the step size. Therefore, in 113 /30s we have 4*113 = 452 hosts. Now we need to find the number of host bits to help us get the subnet mask. Remember the formula 2^y-2. 452=2^y-2…….454=2^y. therefore y is 9 host bits. Remember for hosts we always start counting from the right = N.N.11111110.00000000/23.

By: Anonymous

Anonymous — Tue, 22 Oct 2013 12:45:52 +0000

hi friend this is network address we cant assign it to any host .for this subnet 1st valid ip is .25.last valid .30.31 is brodcast this ip also we cant assign to any host.next subnet is .32.if u hav any query email on..jadhavyogesh150@gmail.com

By: romina

romina — Wed, 02 Oct 2013 06:03:03 +0000

would u please some body explain me the Q.10?

By: Mahad

Mahad — Thu, 26 Sep 2013 18:38:29 +0000

Thanks to Allah and 9tut, I just passed my exam, good luck everyone. I suggest everyone to go thru all the links on this site, close to 50% of question are found here.

By: Anh Tran

Anh Tran — Thu, 26 Sep 2013 16:18:05 +0000

Question 8: 113 point to point links that mean we will need 226 ip address. 226 Subnet mask will be: 32 – 9 = 23 => D the correct.

By: lukin

lukin — Tue, 24 Sep 2013 06:10:07 +0000

I will speak in spanish

tienes la red 172.16.1.0 to 172.16.13.0
el lugar de bit para 172.16.13.0 es 172.16.00001101.0
tomas los primero 4 bit del tercer octeto y son los que quedaran fijos.
172.16.00000001.0
172.16.00000010.0
172.16.00000011.0
hasta
172.16.00001101.0
255.255.11110000.00000000/20
si pones atencion los primeros 4 bit del 3 octeto no se movieron por lo cual debes de ser /20

By: ndane

ndane — Tue, 17 Sep 2013 14:24:22 +0000

q3 can someone explain to me please

By: jasostrong

jasostrong — Wed, 11 Sep 2013 13:40:19 +0000

Thanks 9tut. I made it. I passed my CCNA 200-120 today. The sim is Access-list 1 , Access-list 2 & EIGRP. A lot of new questions like Netflow, Syslog, SNMP, VRRP, and GLBP.

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By: help!!

help!! — Sun, 04 Aug 2013 11:14:52 +0000

@ABP: Use summarized route from phoenix route… hope this helps..

By: ABP

ABP — Mon, 29 Jul 2013 15:50:17 +0000

can any one explain me the question 2

By: Bill_Ado

Bill_Ado — Wed, 24 Jul 2013 21:31:27 +0000

@Nourhan: host B has a valid IP, from the IP 192.168.1.111, we have this range:
Subnet 192.168.1.96, broadcast 192.168.1.127 and the IP assigned to hosts should be anything in between and we’ll have for hosts, the range (192.168.1.97-192.168.1.126), so 192.168.1.111 is a valid host address, i hope it makes sens to you

By: Vick

Vick — Sun, 21 Jul 2013 18:07:14 +0000

freinds please help me i wanna take the CCNA EXAM IN COMING DAYS
please send me last dumps in my email
vickakop@gmail.com
thanks in advance

By: Nourhan

Nourhan — Sat, 20 Jul 2013 03:22:13 +0000

host B has broadcast IP also, Am i right ?

By: RBK

RBK — Tue, 09 Jul 2013 14:35:46 +0000

Hey guys The ip addresses are same or not?

By: Ariel

Ariel — Wed, 26 Jun 2013 08:15:07 +0000

Krishna,
they are on the same network: 192.168.1.32 /27
the range of hosts for this network is 192.168.1.33 – 192.168.1.62

By: feroz

feroz — Wed, 19 Jun 2013 12:36:22 +0000

@bam Thanks for great explanation.

By: Krishna

Krishna — Tue, 18 Jun 2013 02:34:10 +0000

For question 7, option A is valid too…..HostA is not on the same subnet as its default gateway.

Why is that?

By: bam

bam — Fri, 14 Jun 2013 17:18:39 +0000

@ Tin for question number 8, use shortcut the question ask 113 point to point links you can only use here is /30 in a point to point links wich 252 (includes = 1 network address, 2 usable ips and 1 broadcast adress) the total is 4. now 113 * 4 = 452 you need 452 ips. what address can support 452 ips? 1111111.1111111.11111110.00000000 = 2e9 = 512. 512 is enough right? so how many bits are there? /23. so the answer is (255.255.254.0 and 255.255.255.252) or (/23 and /30) which is D.

I hope you get my point sorry my english is bad.

By: Tin

Tin — Sat, 08 Jun 2013 19:04:37 +0000

I mean Q8. Not Q10

By: Tin

Tin — Sat, 08 Jun 2013 19:03:40 +0000

Can someone please explain Q10? Thanks in advance.

By: David Okeri

David Okeri — Wed, 05 Jun 2013 10:21:22 +0000

Pliz i will sit for the CCNA exam next month, can someone send me the latest dumps on email: olesimbe@yahoo.com

By: Richard

Richard — Wed, 15 May 2013 04:37:10 +0000

Kurt,
what do you mean by gateway address? there is no router there yet configured. it only talks subnetting and the number of hosts needed.
Gateway address should be on the same subnet of your hosts.

By: Kurt

Kurt — Fri, 10 May 2013 10:26:22 +0000

Q1 Don’t you think the gateway address of each subnet should be considered?

By: Big Dr.

Big Dr. — Sat, 20 Apr 2013 22:56:03 +0000

Q7 WAS ON MY EXAM TODAY

By: Bhargav

Bhargav — Wed, 17 Apr 2013 17:03:12 +0000

any body wants ccna online training and have any doubts regarding concepts contact me bhargav.cisco@gmail.com

By: Ashique

Ashique — Mon, 08 Apr 2013 22:04:28 +0000

For Question-8
First, consider host address. You need fewest host address, so for point-to-point links, it should be 4 (1 network address, 1 broadcast address & 2 link address) and it take 2 bits. Then you consider subnet address. 113 takes 7 bits. So host & subnet takes 2+7=9 bits. So our initial network should be 32 – 9 = 23

By: ALI

ALI — Fri, 05 Apr 2013 15:07:11 +0000

ques-8
i am thinking same as you , please explain if any one think different .
kind regards

By: here to help

here to help — Sun, 17 Mar 2013 06:04:37 +0000

/26 gives 64 hosts (including the network and broadcast)
/20 gives 4096 hosts (including the network and broadcast)
4096/ 64 = 64 (networks)
that’s what i think is the answer for the first part of the question.

By: Need Help!

Need Help! — Sat, 16 Mar 2013 20:52:51 +0000

My teach ask us to solve this question:

How many /26 aligned networks can be aggregated together to give a /20 contiguous network. What class is the network obtained in this way?

/26 gives us 32 ip that would be 6 subnets
/20 give us 16 ip that would be 12 subnets

thats as far as i got. can anyone help

By: jay

jay — Wed, 13 Mar 2013 14:01:33 +0000

have to consider for number of hosts too. Point-to-point links have 2 hosts.

h=2= (2^y) – 2 thus, y =2.

2 zero bits are borrowed, followed by 7 1′s

By: ques-8

ques-8 — Tue, 12 Mar 2013 19:04:42 +0000

Q8..!!
Answer should be B (/25) Coz 7 bit will be off then and 2^7=128>113

Please Explain

By: faaAdo

faaAdo — Tue, 12 Mar 2013 18:36:49 +0000

Q-4..!!
Sorry i got it, they are asking “which subnet”, 56 isn’t the subnet so the only option left is 64

By: faaAdo

faaAdo — Tue, 12 Mar 2013 18:29:44 +0000

Q-4
Why not Option-D…???

By: shshoor

shshoor — Fri, 01 Mar 2013 23:29:46 +0000

ques 8 was in the exam today

By: Jah Rastafarai

Jah Rastafarai — Wed, 27 Feb 2013 22:29:45 +0000

Passed CCNA, question 8 from here.

By: Rooki3

Rooki3 — Thu, 31 Jan 2013 14:49:26 +0000

Valid Que :D

Q:7

By: route summarization

route summarization — Wed, 30 Jan 2013 20:47:48 +0000

@9tut: do we have a route summarization topic here? thanks

By: Nana

Nana — Wed, 30 Jan 2013 20:04:12 +0000

Guys help me with an activation code for visual certexam manager.i cant access vce files.Even the trial version seems fake.

By: dato pantsulaia

dato pantsulaia — Sun, 27 Jan 2013 13:33:46 +0000

MalikCCNA
three web server need to have three valid ip address on the same sabnet or the last three bits in last octet
00000000.00000000.00000000.00000/000 for that the mask will be
11111111.11111111.11111111.11111 000 or 255.255.255.248

By: tamil

tamil — Fri, 18 Jan 2013 04:51:14 +0000

please help me..why we should give same set of network ip address to serial link
this is my email tamilarasiek@gmail.com

By: TDy11

TDy11 — Thu, 13 Dec 2012 20:14:02 +0000

TY 9tut.

Today I have passed the CCNA. (860/825)
50 questions 3 labs (VTP, EIGRP, ACL). 35 from 9tut.

Also thanks a lot Brar and Sekhar (still valid from examcollection)

Ty again 9tut

from here none

By: James

James — Wed, 12 Dec 2012 00:49:24 +0000

“i dont get why B is correct, E makes more sense to me since you are subnetting from a Class A address.”

It’s not the correct answer is D

Look at it this way:
A. 10.10.0.0/18 – Useable range 10.10.192.0 – 10.10.255.255 (around 7936 ip addresses) subnetted with /30 (1984 blocks of 4 addresses)

B. 10.10.0.0/25 – Useable range 10.10.0.128 – 10.10.0.255 (128 addresses) subnetted with /30 (32 blocks of 4 addresses)

B. 10.10.0.0/24 – Useable range 10.10.0.0 – 10.10.0.255 (256 addresses) subnetted with /30 (64 blocks of 4 addresses)

D. 10.10.0.0/23 – Useable range 10.10.0.0 – 10.10.1.255 (around 512 ip addresses) subnetted with /30 (128 blocks of 4 addresses)

E. 10.10.0.0/16 – Useable range 10.10.0.0 – 10.10.255.255 (65,280 ip addresses) subnetted with /30 (16,230 blocks of 4 addresses)

Since you only have a requirement of 113 WAN Links of four addresses each (network, host, host, broadcast)….some under fill the requirements and some are overkill.

By: sally

sally — Sat, 08 Dec 2012 14:11:23 +0000

Can somebody plz explain this question…

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

i dont get why B is correct, E makes more sense to me since you are subnetting from a Class A address.

By: sally

sally — Sat, 08 Dec 2012 14:10:23 +0000

Can somebody plz explain this question…

A. 10.10.0.018 subnetted with mask 255.255.255.252
B. 10.10.0.025 subnetted with mask 255.255.255.252
C. 10.10.0.024 subnetted with mask 255.255.255.252
D. 10.10.0.023 subnetted with mask 255.255.255.252
E. 10.10.0.016 subnetted with mask 255.255.255.252

i dont get why B is correct, E makes more sense to me since you are subnetting from a Class A address.

By: eddy

eddy — Wed, 28 Nov 2012 06:48:14 +0000

hi
ref Q1
Each subnet has 30 hosts /27
should this not be 6 bits? i thought that 30 hosts would = 30+2 (32) which would be 6 bits
any help?

By: rusure

rusure — Wed, 21 Nov 2012 02:14:12 +0000

the third statement is “acces-list 100 permit ip any any”

By: rusure

rusure — Wed, 21 Nov 2012 01:52:16 +0000

In the ACL examples there is one that says
“permit host B from accessing the finance server..”
“deny host B from accessing the OTHER server not the whole network”

Shouldn’t the first request above say DENY host B from accessing the finance server?
Usually you deny something FROM and permit something TO.
Besides that … permitting it to the server useless as the other hosts also have permission
from the third statement.

By: Godjilaye

Godjilaye — Sun, 18 Nov 2012 20:48:59 +0000

Hallo

Plz can someone explain to me why on question 4 answer B is right? and why not D.

thanks

By: Imran

Imran — Sun, 18 Nov 2012 15:37:05 +0000

Hi, Please can some one explained Question no. 3

By: Bite

Bite — Fri, 19 Oct 2012 03:18:43 +0000

@kai

Which three statements correctly describe Network Device A? (Choose three.)

Here the main Idea is

If they are on the same subnet we need not give the IP address. If they are not then we have to give them

Given two address are 10.1.0.36 and 10.1.1.70

So if we take

A.
With a network wide mask of 255.255.255.128, each interface does not require an IP address.

we can see that if we use this mask they are on different subnet so we need ip, so A is incorrect.
B.
With a network wide mask of 255.255.255.128, each interface does require an IP address on a unique IP subnet.
This is correct as we do require IP addresses

C.
With a network wide mask of 255.255.255.0, must be a Layer 2 device for the PCs to communicate with each other.

NO, we need layer 3 as IP address is involved.

D.
With a network wide mask of 255.255.255.0, must be a Layer 3 device for the PCs to communicate with each other.
YES.
E.
With a network wide mask of 255.255.254.0, each interface does not require an IP address.
Yes, as both are on the same subnet

Thanks

By: Bite

Bite — Fri, 19 Oct 2012 02:59:41 +0000

@Q8

Hi All!

This almost wated a lot of time on this so just sharing what I have learnt if this helps!

when we do according to the basics we know we have for 113 links we need is

2*2*2*2*2*2*2=7 bits to be reserved so

we have is some class address where 7 bits have to be reserved.

from the given options below we see them like this

10.10.0.0/23 subnetted with mask 255.255.255.252

so for 255.255.255.252 is nothing but /30 so if we

by increasing given /23 to /30 we have increased address by 30-23=7

which can give 2^7 addresses we need.

say if we take other options we have

10.10.0.0/25 subnetted with mask 255.255.255.252

/25 to /30 we have 30-25=5

2^5=32

but we need 113
Hope this helps

Thanks

By: kai

kai — Sun, 09 Sep 2012 02:29:28 +0000

can somebody help me with this question please? http://www.aiotestking.com/cisco/2012/04/which-three-statements-correctly-describe-network-device-a/

By: sniffer

sniffer — Fri, 07 Sep 2012 23:22:51 +0000

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

@ me and xallax meaning the rite ans is 172.1.4.0/22 not 172.1.0.0/22 as in the dump.
@ me check from ur dump if 172.1.4.0 is part of the ans????

By: 9tut

9tut — Fri, 07 Sep 2012 10:53:16 +0000

@Tedy_bear: It’s my mistake. Thanks for your detection, I fixed it!

By: Tedy_bear

Tedy_bear — Fri, 07 Sep 2012 09:22:37 +0000

@9tut Please check for typo – Right Menu URLs showing the same Questions content
http://www.9tut.com/ccna-subnetting-questions-3
http://www.9tut.com/ccna-subnetting-questions-4

By: xallax

xallax — Sat, 01 Sep 2012 13:11:23 +0000

@malikccna
the smallest subnet to fit 3 servers is the one from option C

By: MalikCCNA

MalikCCNA — Sat, 01 Sep 2012 05:07:34 +0000

You have been asked to come up with a subnet mask that will allow all three web seivers to be on the same network while proving the maximum number of subnets. Which network address and subnet mask meet this requirement?

A.
192.168.252.8 255.255.255.252

B.
192.168.252.16 255.255.255.252

C.
192.168.252.8 255.255.255.248

D.
192.168.252.0 255.255.255.252

E.
192.168.252.16 255.255.255.240
——————————————–

Why the answer is C..?
Can anyone explain please..?

By: jacsatx

jacsatx — Wed, 29 Aug 2012 21:55:37 +0000

no dumps needed on subetting..either u know it or not….907!!..thx 9tut!!

By: atman

atman — Wed, 29 Aug 2012 03:48:33 +0000

Can anyone mail me dumps please? atman.trivedi@hotmail.com Many thanks

By: latha

latha — Sun, 12 Aug 2012 08:42:49 +0000

than u very much

By: khurrum

khurrum — Wed, 01 Aug 2012 13:47:10 +0000

hi, could someone please simplify Q3 for me please?

By: CCNA as of 8/3/12

CCNA as of 8/3/12 — Sat, 21 Jul 2012 16:21:53 +0000

@ Anonymous

The Network 172.25.0.0 has been divided into 8 equal subnets. Which of the following IP address can be assigned to the hosts in the 3rd subnet if the IP subnet-zero command is configured on the router? Choose 3 correct answers.
1. 172.25.78.243
2. 172.25.98.16
3. 172.25.72.0
4. 172.25.94.255
5. 172.25.96.17
6 172.25.100.16
correct ans according to dump is
1 3 and 4.

If you have 8 equal subnets you are looking at 32 hosts per subnet. Your subnets would be:

172.25.0.x
172.25.32.x
172.25.64.x
172.25.96.x

This question is specifically looking for hosts in the 3rd subnet. So we would be interested in:

172.25.64.1 – 172.25.64.95.254

Therefore, the correct answers given the choices are 1, 3 and 4

By: Rajan

Rajan — Wed, 18 Jul 2012 05:36:37 +0000

I find Wendell Odom’s product to be more inivttiue and easier to just jump right in. In my opinion, the design process that goes into designing any network simulation software, is similar to designing a game. Some games require you to learn the controls, before you can really get started, while others are so well thought out that you can well, just jump right in. However, if anyone has any plans to progress beyond the CCNA (eg. CCNP, etc.), I would recommend taking a look at the other two simulators. When I first purchased RouterSim’s Network Visualizer, it was still a buggy product and prone to crashes. Even so, it was the only worthwhile game in town, at the time (2004). I never got the opportunity to finish the labs, as I was called to active duty and served over seas for almost two years. RouterSim’s Network Visualizer has a particularly strict software license. You only get one, and in order to install it on another machine, you have to send your activation key back to the company (online, via the software itself). Once the license is back with the company, you can then install the software again on another computer, and then activate the product by downloading activation key from them once again. Ever served in the military? Well when you get those orders, you go. There is not a lot of time. You may have a few days to settle any pending business, like saying goodbye to your family and friends, making sure your will is in order, arrange for your bills to be paid, and so on. So transferring a software license wasn’t on the top of my list. My software was originally installed on a desktop and I was not about to lug that (and the monitor) with me in my ruck. By the time I came back, the CCNA exam had changed so much that my simulator was, yes outdated. I tried explaining my situation to the nice folks at RouterSim, but I was informed nicely that I can just purchase the upgrade. Remembering my experiences with the software crashes, I declined and signed up to take a CCNA cram class instead. Big mistake. For a newbie, cram courses don’t work. Eight days (8 hours a day) of instruction will not teach you the operations of a network and how Routers and Switches come into play so much so that you can pass the CCNA. Much less, get you so familiar with the input commands that you can use on the job. In my class, I had an Egyptian instructor who may be very knowledgeable, but help me if I can understand a darn thing that he was saying. I just couldn’t get pass the accent. The school did however, introduced me to Boson’s NetSim, a far superior product than my older version of RouterSim’s Network Visualizer. Oh yes, if you decide on taking a class, make sure that they give you access to ACTUAL Cisco equipment! They didn’t technically lie to me however, the school did have Cisco equipment, but that was a very brief show and tell. We primarily were working with the NetSim software. There are several versions now CCENT, CCNA, CCNP, etc. They also sell videos and mp3 s, besides the simulation software. Cisco seems to be endorsing them, you can find links to Boson from the Cisco website. To briefly describe the product, you basically have a network simulation software package along with a couple of pdf lab manuals (they are compressed and will appear once you’ve installed the software). I would print out the manuals and follow through the lessons, chapter by chapter. So why do I prefer Wendell Odom’s Network Simulator over the others? One, you get 4 licenses. Yes, you can install your product on FOUR separate computers. Two, the price. Wendell Odom’s Network Simulator is under $100 (Amazon price, retail is $149). RouterSim’s CCNA Network Visualizer is $229.00. And Boson’s NetSim for CCNA is $249. And third, the pdf step-by-step lessons on Wendell Odom’s Network Simulator pops up as you activate the corresponding lesson simulation. Something that I can really appreciate, as I will be traveling a lot on my new job, and don’t want to carry a hard copy lesson book with me, nor scroll through the pages to the lesson I want. Dual monitors would serve you well here. It really is just simpler to use. You are presented with a lesson menu on the right side. A diagram of your network is on the top left, and the command line section (where you type) is below that. Click on the computer icon on your virtual network, and on the command line section you’re looking at the CLI interpretation of what you would see if you were logged on to that computer. The same goes for the switches and the routers. Neat. Do bear in mind however, that network simulators are

By: singh

singh — Mon, 16 Jul 2012 19:48:25 +0000

If they given the mask /27 then the subnet 0,32,64,96————-so on.The 3rd subnet is 64 and ip add. range between 64 to 95. That by the correct ans is 1 3 and 4. thanks

By: Anonymous

Anonymous — Mon, 16 Jul 2012 12:49:47 +0000

1. 172.25.78.243

2. 172.25.98.16

3. 172.25.72.0

4. 172.25.94.255

5. 172.25.96.17

6 172.25.100.16

correct ans according to dump is

1 3 and 4.

9TUT could anyone explain how this is calculated?

regards,
Rutz

By: 9tut

9tut — Sun, 15 Jul 2012 23:14:20 +0000

@youngj: Please read my Subnetting tutorial: http://www.9tut.com/subnetting-tutorial

By: youngj

youngj — Sun, 15 Jul 2012 18:53:14 +0000

this subnetting is really confusing becuz everyone does it their way……….hmmmmmm i am writing my test at the end of this month ……any good help regarding subnetting ,vlsm and summarization questions ???????????i wanna get them done easily …wts the best way /// plz help

By: dinky

dinky — Thu, 12 Jul 2012 09:17:27 +0000

hi sumit..

if you calculate the range or block size it will come out to be 64 and there will be 4 subnets in this case. well first subnet is 192.168.1.0 and second is 192.168.1.64
first three networks in the diagram are from first range so it has to be from 192.168.1.64/26

hope u got it.

By: Sumit

Sumit — Thu, 12 Jul 2012 07:46:01 +0000

hello,
i got a problem with Q4. i want to know how option B is the correct answer when option D is also with /26 ?

By: dinky

dinky — Sun, 08 Jul 2012 13:42:55 +0000

Hello all,

I got a prob with que no 2. Please brief me how we move 2 bits to left from 24 to 22?

Thanks

By: Chris

Chris — Fri, 06 Jul 2012 13:10:43 +0000

Q8,

I found out how it is answer D. What you have to do is look at the subnet mask and which is /30 and then to find the difference between each of the networks which gives you:

A – 12
B – 5
C – 6
D – 7
E – 14

So you then work out how many of subnets for this number of bits, 7 bits gives you 128 which is the best fit for the requirement in the question of 113 networks, with 2 hosts each (point-to-point). Hope this makes more sense now.

By: Chris

Chris — Tue, 03 Jul 2012 14:07:57 +0000

For Q8 I put answer D, but when I was working it out I made the mistake of counting the subnets as if it was a class B address.

As it is class A then Kurt is right that the amount of subnets will be these huge figures given so the question could be flawed.

Is it that you are discarding the first two octets because 10.10.0.0 appears to be the network ID you are starting with so you mask out the first two octets and count from there 2, 4, 8, 16, 32, 64, 128 which fits in 113 subnets therefore it is /23. You would then have network IDs of 10.10.0.0, 10.10.2.0, 10.10.4.0, 10.10.6.0, etc. ?

By: God's Grace

God's Grace — Thu, 21 Jun 2012 00:29:51 +0000

I passed my ccna exam today Praise be to God! Thank you Jesus! and thanks to 9TUT for the tutorials and explanations, great site and thanks to xallax for your explanations to questions and thanks to http://www.examcollection.com for the dumps. Pls guys lets donate and help to keep this site up!

48 ques for exams including 3 simulation, I had EIGRP, Acesslist2 and VTP. Make sure the practice the simulation, use packet tracer or gns3. Best wishes to all!

By: dougenius

dougenius — Wed, 20 Jun 2012 12:40:59 +0000

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By: kurt

kurt — Thu, 14 Jun 2012 18:14:38 +0000

/15 = 128 131,070 = 131,198
/16 = 256 65,534 = 65,790
/17 = 512 32,766 = 33,278
/18 = 1,024 16,382 = 17,406

I thought /15 -/17 would be better but /18 appears to be the best one. It wastes the least amount of Host & Subnets.

/20 would be the best (4096 + 4094 = 8190)

Please someone point out what I am doing wrong. I think I am correct here.

By: kurt

kurt — Thu, 14 Jun 2012 18:05:22 +0000

NOTE: 113 point-to-point links, thus this means subnet addresses, Not Hosts.

By: kurt

kurt — Thu, 14 Jun 2012 18:01:26 +0000

Q8 is still confusing after reading all of these answers and explanations.

And moncho was using ClassB subnets, these are Class A, not B!

Subnets Hosts Total
A. 10.10.0.0/18 = 1,024 16,382 = 17,406
B. 10.10.0.0/25 = 131,072 126 = 131,198
C. 10.10.0.0/24 = 65,536 254 = 65,790
D. 10.10.0.0/23 = 32,768 510 = 33,278
E. 10.10.0.0/16 = 256 65,534 = 65,790

If you want the least amount of Host & Subnets, then A would be your best bet, but all of these are terrible answers.

Please someone let me know what I am missing here.