Comments on: CCNA – Subnetting Questions 3

By: Anonymous

Anonymous — Wed, 06 Aug 2014 04:17:10 +0000

brilliantly presented and explained.

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By: Anonymous

Anonymous — Sat, 09 Nov 2013 08:42:10 +0000

There r no subnetting questions in the exam.THJe paper consists of more questions from
OSPF.

By: Anonymous

Anonymous — Sat, 09 Nov 2013 08:40:37 +0000

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By: kavi

kavi — Mon, 30 Sep 2013 17:50:08 +0000

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By: hh86

hh86 — Tue, 24 Sep 2013 15:24:17 +0000

q 3, 10 ,11 in exam yesterday

By: lukin

lukin — Tue, 24 Sep 2013 05:31:43 +0000

/25=128, /26=192, /27=224 , “/28=240″, /29=248, /30=252

/28=240= hop is (16) is 0,16,32,48,64,80,96,112,128,144…….
and

/25=128 hop is (128) is 0 to 127 and 128 to 255 the ip assigned is 192.0.2.24 and 192.0.2.100

By: Waymond

Waymond — Wed, 18 Sep 2013 00:22:48 +0000

I understand Q1 but what would be a fast way of solving it? I had to calculate everything out.

By: jasostrong

jasostrong — Wed, 11 Sep 2013 13:40:11 +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.

By: csj_0914

csj_0914 — Sat, 07 Sep 2013 12:20:08 +0000

Q6. 115.64.4.0/22
INCREMENT : 4
NETWORK: 115.64.4.0
BROADCAST: 115.64.7.255

Below choices are within this range of network 115.64.4.0/22
B. 115.64.7.64
C. 115.64.6.255
E. 115.64.5.128

love subnetting.

By: padmini

padmini — Thu, 22 Aug 2013 06:27:06 +0000

Q7 B is not a correct answer because the ip address falls under different subnetwork.
For the subnetwork 192.1.144.0/20 , the valid ip address are 192.1.144.1 to 192.1.159.254.
and 192.1.159.255 is the broadcast address.

By: alexio

alexio — Thu, 22 Aug 2013 01:29:59 +0000

Will be grateful if someone can explain Q7 especially why B is not a correct answer

By: :-)

:-) — Mon, 19 Aug 2013 15:54:08 +0000

I think there is an easy solution for Q11
A-Public IP
C-Loop Back Address
D-Multicast address
E-Public IP
we cant assign these IP’s to hosts !

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By: Ahmad

Ahmad — Mon, 12 Aug 2013 14:38:04 +0000

For question 4, the work station’s IP 131.1.123.24/27, means 32-27= 5 right most bits can be used for host processing. So the magic number is 2^5=32(or 256-224=32 from the chart). Thus the IP range will be 131.1.123.0—131.1.123.31
131.1.123.32—131.1.123.63
131.1.123.64—131.1.123.95
131.1.123.96—-etc.
Now see, the IP address of the host is in range 131.1.123.0—131.1.123.31, where as the GW is within the range 131.1.123.32—131.1.123.63, means different network. Am I clear?

By: kufkov

kufkov — Sun, 11 Aug 2013 16:37:17 +0000

someone should please explain question 4 and 7 for me.I will very grateful

By: Ahmad

Ahmad — Sun, 11 Aug 2013 13:49:34 +0000

Q 7: why answer B not correct? Plz let me know.

By: lala

lala — Sun, 28 Jul 2013 13:02:55 +0000

Hi i can’t understand at Question 8:

Why isn’t the link between core and sales in the same subnet as SALES .

The range for sales is 172.16.3.64- 172.16.3.95

And the link is 172.16.3.8/30 which is not included in that /27

Or the link between Core and Admin in the same Subnet as Admin

The range for Admin is 172.16.3.96-172.16.3.127

And the link is 172.16.3.12/30 which is not included in that /27

Also for Marketing

Why none of the serial links don’t belong to the department’s /27

From what I can see they are all included in a /28 (4 subnets of /30)

and they range from

172.16.3.0-172.16.3.15 which is has nothing to do with the other /27 .

The same is for Network A which ranges from

172.16.3.128-172.16.3.191 which has nothing to do with the links /30

Shouldn’t the /30 be in the same subnet as their department, as their /27 ?

By: Bill_Ado

Bill_Ado — Wed, 24 Jul 2013 21:05:05 +0000

@Anonymous, July 18th, just to complete my friend GLENN, and remember also that here we are working in the third octect, so all our networks are changing in the third octet, for example:
191.168.0.0, 191.168.2.0,……….191.168.10.0 and the range of usable hosts for this subnet is
(191.168.10.1-191.168.11.254), and 191.168.10.2 is included in this range. broacast 191.168.11.255, and we can not assign network and broadcast addresses to devices, remember?

By: Vick

Vick — Sun, 21 Jul 2013 17:08:09 +0000

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By: GLENN

GLENN — Fri, 19 Jul 2013 10:22:31 +0000

Hi Anonymous / Edwinklain

With 191.168.10.2/23,

the network address is 191.168.10.0/23
the host address range is from 191.168.10.1 to 192.168.11.254
So B is correct.

192.168.10.2

By: Anonymous

Anonymous — Thu, 18 Jul 2013 09:31:10 +0000

in question 11 how can the answer include B ie B. 191.168.10.2/23.I thought this ip is a network address.
increment = 2
191.168.10.2/23 it should be the network address of subnet 2
191.168.10.0 -191.168.10.1

kindly somebody explain to me.
thanks

Edwinklain

By: lumpachi

lumpachi — Wed, 03 Jul 2013 14:12:43 +0000

hello Ronin,

network A needs 120 host, so a subnet mask of /25 is needed to accommodate this.
(2^7)-2 (network address and broadcast address cant be used) = 126 (it can accommodate until 126 hosts.
32 – 7 (bits fr. the above formula) = /25 (Letter A and C are wrong since it uses /26 subnet mask which can only accommodate 62 hosts)

and since, subnet ip zero command is configured, Letter D – 172.16.3.0/30 can be used as its network address. (Letter E and F are wrong because the subnet belongs to Marketing and Admin subnet respectively)

By: Mapampola

Mapampola — Wed, 03 Jul 2013 11:25:06 +0000

Hi Budoy

Note that the increment is 4 on the 3rd octet. The subnet is 115.64.4.0 – 115.64.7.255. Any address in this subnet is a valid IP

By: budoy

budoy — Wed, 03 Jul 2013 09:28:30 +0000

please explain question 8 more clear! PLEASE! why is subnet /30 on subnet /27

By: budoy

budoy — Wed, 03 Jul 2013 09:27:44 +0000

please explain question 6 more clear!

By: Ronin

Ronin — Tue, 02 Jul 2013 04:12:32 +0000

can question 8 be explained please?

By: Zipek

Zipek — Wed, 26 Jun 2013 00:24:40 +0000

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By: Lu

Lu — Mon, 10 Jun 2013 11:18:53 +0000

because D incorrect?

By: Lu

Lu — Mon, 10 Jun 2013 11:18:04 +0000

you can explain q5??? thanks…

By: samar

samar — Sat, 08 Jun 2013 11:37:18 +0000

can someone explain question 7 please?

By: David Okeri

David Okeri — Wed, 05 Jun 2013 09:25:57 +0000

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

By: Drexx

Drexx — Tue, 04 Jun 2013 22:50:26 +0000

Karol thanks a lot for your explanation about Q7. Bless you.

By: @chatembwa

@chatembwa — Mon, 20 May 2013 14:55:49 +0000

may someone explain explain to me, the logic behind Q3

By: saurabh kukreja

saurabh kukreja — Thu, 09 May 2013 15:46:53 +0000

D,E are the correct answers

By: Diehko

Diehko — Sat, 27 Apr 2013 15:38:05 +0000

I think all the questions are right. You must sduty a little more and then you can discuss about them. Thanks 9tut, I’ll make a contribution when I can.

By: dev

dev — Fri, 19 Apr 2013 21:01:20 +0000

@exam taker

You are attacking the question as if you are completely dumping the /23 mask and then going to an entirely different /27 mask. However, the you are taking the already existing /23 mask: 11111111.11111111.11111110.00000000 and turning it into…
11111111.11111111.11111111.11100000 (/27)

Therefor you are adding 4 bits to the network portion of the address. 2^4=16 subnets
the 5 bits that remain off represent the hosts: 2^5-2 = 30 hosts

So, 16 subnets and 30 hosts.

Good question though, that is tricky.

By: examtaker

examtaker — Tue, 16 Apr 2013 15:10:09 +0000

Question 12

How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?

A. 8 subnets, 31 hosts
B. 8 subnets, 32 hosts
C. 16 subnets, 30 hosts
D. 16 subnets, 32 hosts
E. A Class B address cant be subnetted into the fourth octet.

only 16 subnets gained? i can’t believe it…
i thought
/23–>has 7 bits on–>2^7=128 subnets
/27–>has 11 bits on–>2^11=2048 subnets
#of subnets gained=2048-128=1920

By: SAEED

SAEED — Wed, 10 Apr 2013 18:59:14 +0000

Q12 PLEASE CAN ANY ONE EXPALIN IT BREIFLYY ????

By: khan

khan — Sat, 23 Mar 2013 12:09:09 +0000

plz anyone can help me 1st question ?

By: Ahmad

Ahmad — Wed, 13 Mar 2013 16:09:34 +0000

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By: fantagista

fantagista — Wed, 13 Mar 2013 01:57:28 +0000

It’s very easy~~~~i want hardmode

By: ben

ben — Mon, 11 Mar 2013 15:25:51 +0000

@jimmy
192.1.144.0/20 –> /20 means increment of 16 and belong to 3rd octet. so, 16, 32, 48, 64….., since the choices is higher number, you can easy jump up to 128, 144, 160.. then as you can see network 192.1.144.0/20, what usable ip address belongs to this network address?, looking back to the options, you can correctly select the option A. 192.1.159.2 and D. 192.1.151.254. take note that were looking at the 3rd octet. next network address is 192.1.160.0/20.

By: Anonymous

Anonymous — Thu, 07 Mar 2013 19:34:06 +0000

I had question 4 on the ICND 2 exam today.

By: jimmy

jimmy — Thu, 07 Mar 2013 15:49:04 +0000

plz explain

QUESTION 7

By: MH

MH — Mon, 04 Mar 2013 06:20:03 +0000

@Ali
/22 means Block of 4 i.e. 0,4,8,12 …….
Given Address is 115.64.4.0 therefore It lies between 4 and 8 . Exclude network and broadcast address i.e. 115.64.4.0 & 115.64.7.255 thus all address in between this range are valid.

By: shshoor

shshoor — Fri, 01 Mar 2013 23:29:05 +0000

ques no 3 and 11 was in the exam today

By: ali

ali — Thu, 28 Feb 2013 07:39:25 +0000

somebody please explain to me q6 ??????

By: Jah Rastafarai

Jah Rastafarai — Wed, 27 Feb 2013 22:28:47 +0000

Passed CCNA, question 11 from here.

By: brent

brent — Sun, 24 Feb 2013 22:09:43 +0000

@mark Terrell /23 means we’ve got 23 bits turned on as 1111 1111.1111 1111.1111 1110.0000 0000 so we’re left with 9 bits for hosts 2 ^ 9 = 512 divided by 32 gives us 16 subnets. /27 it’s 32 hosts minus 2 = 30 hosts so the correct answer should be 16 subnets and 30 hosts. i hope that helps.

By: Daniel

Daniel — Wed, 20 Feb 2013 14:42:57 +0000

I’ve just look better. My mistake about Q7

By: Daniel

Daniel — Wed, 20 Feb 2013 14:39:46 +0000

D and E are the correct answers.
Increment 16 so the Network Address will be: 192.1.140.0; Broadcast: 192.1.155.255. The range is between 192.1.140.1 to 192.1.155.254 so D and E are the correct answers.

Am I wrong? please correct me if so. Ty

By: saro

saro — Wed, 20 Feb 2013 05:45:10 +0000

cn anyone xplain q.no 6???????????

By: Mark Terrell

Mark Terrell — Tue, 19 Feb 2013 19:17:23 +0000

Excellent site–keep the questions coming!

By: Mark Terrell

Mark Terrell — Tue, 19 Feb 2013 15:16:21 +0000

Question 12-is wrong. You didn’t add your bits right. The class B/23 gives you-2^7 or 128 subnets. By taking this class B to a /27, you would add the 4 bits to your 2^7 giving you 2^11 or 2048 subnets.

By: Gauri

Gauri — Sun, 17 Feb 2013 12:32:02 +0000

In question no 7:
the subnet we are looking at is : 192.1.144.0-192.1.159.255
where, 192.1.159.255 is the broadcast address
So the valid hosts are 192.1.144.1 – 192.1.159.254
So, why is 192.1.159.2 not selected over the other two options?

kindly correct me if I’m missing any points.

By: N k Tiwari

N k Tiwari — Wed, 06 Feb 2013 07:46:36 +0000

Thanks a lot. So easy explain.

By: mmr

mmr — Thu, 31 Jan 2013 10:07:34 +0000

ans. of the ques. no.3 is not clear.

By: George Sam

George Sam — Sun, 20 Jan 2013 10:08:45 +0000

JJ
In question4 Server A and Server B in /27 network. Valid host range in Network A is 131.1.123.1 to 131.1.123.30 Valid host range in network B is 131.1.123.33 to 131.1.123.62
Here the gateway provided for both network is 131.1.123.33 which comes in Network B subnet not in A. Gateway of Server A is not in the same subnet.

By: JJ

JJ — Sun, 20 Jan 2013 09:08:25 +0000

can anyone explain Q4 please ?

By: Raman

Raman — Sun, 20 Jan 2013 01:50:52 +0000

@Karol

Why u subtract 1 from 16 in the solution of ques:7 {192.1.159.0 (144+16-1)}

By: chuckles

chuckles — Sat, 19 Jan 2013 12:11:09 +0000

can any ccna professional give me his email so that i can ask him question that i get stuck at in dumps

By: Muhammad Arshad

Muhammad Arshad — Sun, 13 Jan 2013 13:48:23 +0000

@Chuckles::: Q# 5 Explanation::::

192.168.20.19/28 is an IP address belongs to 192.168.20.16 Network.
as you Know /28 means Increament of 16 so the next Network address is 192.168.20.32
so the valid host are b/w them which are 192.168.20.17 To 192.168.20.30 while the 192.168.20.31 is the broadcast address.
According to Options
A. 192.168.20.29 (belongs to the same range so its correct)
B. 192.168.20.16 ( is the network Address so we will not opt for it)
C. 192.168.20.17 ( is the First Address in the range so Correct)
D. 192.168.20.31 ( is the Broadcast Address and is not valid so incorrect)
E. 192.168.20.0 ( is a net work Address and valid hosts are (1-14)
Regards.

By: Muhammad Arshad

Muhammad Arshad — Sun, 13 Jan 2013 13:31:05 +0000

@mutazzzzzzzzzzzz::

You will pass but not understand. I mean first Study the CCNA Syllabus and i will suggest Todd Lammele’s 7th Edition a very easy to understand book with hands on Exercises and Q&A’s.

after completing the Book then use dumps to practice and you will pass CCNA.

By: sahar

sahar — Fri, 04 Jan 2013 20:53:43 +0000

karol
thanks a lot

By: mutazzzzzzzzzzzz

mutazzzzzzzzzzzz — Fri, 04 Jan 2013 20:47:15 +0000

if i learn dumbs so welll i will passsssssssss?

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mutazzzzzzzzzzzz — Fri, 04 Jan 2013 20:46:32 +0000

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By: Karol

Karol — Fri, 04 Jan 2013 19:56:04 +0000

sahar
Answer Q7, the network is 192.1.144.0/20 so 192.1.144.0 is network address and broadcast is 192.1.159.0 (144+16-1). 16 because of mask 4 bits on third octet. The host are between 192.1.144.1 to 192.1.159.255.

By: sahar

sahar — Fri, 04 Jan 2013 10:57:29 +0000

Can anyone explain Q7 plz?

By: Eric K.

Eric K. — Thu, 27 Dec 2012 14:30:44 +0000

Which subnetting section should I see in ICDN1?

By: chuckles

chuckles — Tue, 25 Dec 2012 14:01:45 +0000

can someone explain Q5 please , isnt the network address is 192.168.20.19 and the broadcast address is 192.168.20.50 , answer c dont seem right to me

By: makoi

makoi — Sun, 23 Dec 2012 12:18:04 +0000

how did arrive to /22 can someone pls explain thanks

By: Kaganskan

Kaganskan — Sun, 23 Dec 2012 03:42:34 +0000

Thank you dhina where did the 100 host came from the question?
is it from 192.0.2.100/28?

By: dhina

dhina — Sat, 22 Dec 2012 17:35:48 +0000

Hi Kagan
How many bits we need to get 100 host 2^7=128 right.. so save 7 bit for host balance 1 bit borrow from host portion to network. so we will get /25…U got it..

By: Kaganskan

Kaganskan — Sat, 22 Dec 2012 13:50:34 +0000

Can anyone explain briefly Q1?
I’m a little bit confused about /25 answer

By: 9tut

9tut — Sun, 16 Dec 2012 00:02:17 +0000

@Jamerican: Yes, the next network will be 96 so the current network is from 172.31.80.5 to 172.31.95.255. I didn’t say anything about “92″ in Q9.

Increment: 16
Network address: 172.31.80.0
Broadcast address: 172.31.95.255

By: Jamerican

Jamerican — Sat, 15 Dec 2012 21:27:03 +0000

9tut, Q9 does not make sense. It specifically ask “Which two subnetworks would be included in the summarized address of 172.31.80.0/20?” If you are incrementing by 16 the next network will be 96 not 92. By arriving at 92 you would have to be incrementing by 4 which is a /22 but that is not what the question is asking. Could you please breakdown the question for me? Thanks

By: TDy11

TDy11 — Thu, 13 Dec 2012 20:13:34 +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 Q 5

By: Kahlan

Kahlan — Sat, 08 Dec 2012 14:35:26 +0000

@sally

first, the right topic is here : http://www.9tut.com/ccna-subnetting-questions-4
it’s the question 8 of subnetting question 4.

And 9tut explained it :
“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)"

By: sally

sally — Sat, 08 Dec 2012 14:01:41 +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.