Comments on: New CCNA – Subnetting

By: Anonymous

Anonymous — Sun, 31 May 2015 01:16:27 +0000

Regarding Q9, the mask of answer D is wrong.
D. 10.10.0.0/23 subnetted with mask 255.255.255.252

Mask of 10.10.0.0/23 should be 255.255.254.0

By: Anonymous

Anonymous — Fri, 29 May 2015 02:11:55 +0000

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By: Q9

Q9 — Wed, 27 May 2015 04:20:01 +0000

/30 is used for p-t-p because u only need 2 ip address from each end of the point to point link.

30-7= 23
30 (1s)
7(0s)

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By: Anonymous

Anonymous — Sun, 24 May 2015 21:23:25 +0000

@9tut please explain Q9 in more detail
Many thanks

By: Anonymous

Anonymous — Sun, 24 May 2015 21:11:13 +0000

Q9.
easy way to understand this trick question. It says point-to-point links, 113 hosts instead of saying subnets. If you have to borrow 0s from the 1s, so you are getting hosts.
2^7=128, then you have to borrow 7 bits from the mask /30. You can do 30-7= 23 or /23. So thats finish.

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By: Pascal

Pascal — Sun, 10 May 2015 10:42:51 +0000

Q4 answer is not clear at all :
the answer should be 255.255.255.0 :
eight /27 subnets = one /24

It’s a classful /24 but what is the problem ?

By: Pascal

Pascal — Sun, 10 May 2015 10:40:33 +0000

Q4 is not cela

By: Pascal

Pascal — Sun, 10 May 2015 10:30:47 +0000

Q2 : is the following explanation right ?
“Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.”

According to me, the ip subnet-zero refers to classfull networks and as 172.16 is a class B, its subnet-zero shoud be : 172.16.0.0 /n (n> or =16) and then 172.16.3.0/30 is not a subnet zero.

The no ip subnet-zero could have been set, it would not have changed the result.

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By: Q6

Q6 — Tue, 28 Apr 2015 16:20:14 +0000

Hi,
Can someone explain q6 please?

By: Tomas

Tomas — Tue, 21 Apr 2015 10:25:54 +0000

to Martin:
You need at least 26 hosts.

However option B gives you this mask: 255.255.255.252
Written in binary: 11111111.11111111.11111111.11111100
There are only 2 zeros
2^2 = 4 – you may think you could address 4 hosts:
x.x.x.x00
x.x.x.x01
x.x.x.x10
x.x.x.x11

! HOWEVER !

IP address with all 0s is the NETWORK ADDRESS and can’t be used. Therefore x.x.x.x00 can’t be used to address host in LAN.

IP address with all 1s is called BROADCAST and also can’t be used. Therefore x.x.x.x11 can’t be used to address host in LAN.

THEREFORE, Everytime you are subnetting, you must subtract 2 (1 for network IP +1 for broadcast).

Therefore option B is not correct: 2^2 = 4-2 = ONLY 2 hosts. However you require 25 hosts.
2<25.

I hope, all is clear now.

By: Martin

Martin — Thu, 16 Apr 2015 08:07:13 +0000

I think bit “1″ represent network. bit ’0′ represent host ..

By: Martin

Martin — Thu, 16 Apr 2015 07:14:22 +0000

Q4: hi anyone can explain me, I don’t understand below explanation …2^2-2=2..
For answer B: 252 = 1111 1100 -> with this subnet mask we can subnet up to 26 = 64 subnets but only 2^2 – 2 = 2 hosts per subnet -> B is not correct.

By: Wasim

Wasim — Fri, 10 Apr 2015 08:55:33 +0000

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By: Ted123

Ted123 — Mon, 30 Mar 2015 12:14:24 +0000

Passed the exam with flying colours 9tut does well

EIGRP 212
ACL MOD 3
VLAN Trunking
ACL
STP
Switching
Netflow
SNMP
Subnetting

Basically all the questions where here besides 4 new questions. You guys will be fine just practise these questions to get a good understanding of what CCNA is about. Make sure you understand what each section is about e.g. frame-relay, WAN, etc etc.

Because at the end of the day you can pass the exam but wont understand j@ck Sh!t about what networking is all about.

By: Shahin

Shahin — Sat, 28 Mar 2015 18:52:32 +0000

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By: PAUL

PAUL — Wed, 25 Mar 2015 17:13:47 +0000

Please explain Q5 :(

By: nata

nata — Wed, 25 Mar 2015 08:15:39 +0000

could you pls justify ans for Q No 10 .

By: mike

mike — Wed, 25 Mar 2015 06:46:30 +0000

Can you please sent the dumb to email michaelyoussef888@gmail.com thanks

By: Apush

Apush — Wed, 18 Mar 2015 12:01:11 +0000

Passed my CCNA exam today (18th Mar)… Q8 in exam

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By: santosh

santosh — Mon, 02 Mar 2015 01:51:37 +0000

Q6. I guess all of the answer are incorrect, the summarization should be 172.16.32.0/19. Guys, what u came up with this question? Help???

By: illbwolf

illbwolf — Thu, 19 Feb 2015 23:35:25 +0000

q3,8,9

By: Hasan

Hasan — Tue, 17 Feb 2015 10:36:07 +0000

i passed CNNA tody and Q6 was inclouded

By: Che

Che — Mon, 16 Feb 2015 05:12:55 +0000

Q6 On 15th Feb.

By: sriivas

sriivas — Wed, 11 Feb 2015 18:13:43 +0000

alaaah very easy questions

By: droxz

droxz — Wed, 11 Feb 2015 00:13:33 +0000

Q1 is B

you have 310 hosts:

255.255.255.0 = can accomodate 254 hosts < 310
255.255.254.0 = can accomodate 510 hosts
255.255.252.0 = can accomodate 1022 hosts

since the question asked for the same subnet with least amount of wated address, so B is your best bet.

By: altaf

altaf — Tue, 10 Feb 2015 15:14:25 +0000

Q6 in my exam paper today got 1000 marks

By: MikeTheMan

MikeTheMan — Tue, 10 Feb 2015 14:58:33 +0000

Q1 .. Answer should be C not B

310 hosts 255.255.252.0

By: Yannick

Yannick — Tue, 10 Feb 2015 14:13:47 +0000

Question 4

Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?

A. 0.0.0.240
B. 255.255.255.252
C. 255.255.255.0
D. 255.255.255.224
E. 255.255.255.240

In my opinion answer is C and not D.

Question: :

Which subnet mask would be appropriate (X) to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?

Explanation:
26 Hosts : /27 ( 255.255.255.224)
In a /26 there is two /27
In a /25 there is four /27
In a /24 there is eight /27

So the answer is /24 = 255.255.255.0 => Answer C

By: lyra

lyra — Tue, 03 Feb 2015 21:04:36 +0000

Faizan,
For Q7; it asked for as many as 30 hosts which means we need 5 bits i.e 2^5-2 = 30 hosts
10.188.31.0/23
/23 = 255.255.254.0
9 bits available to play with X.XXXXXXXX
We need 5 bits reserved it from Right to Left(for Host) i.e 1.11100000 rest goes to network
so new Mask will be 255.255.255.224 i.e /27

Mask Above /27 have less then 30 hosts
Mask below /27 have more then 30 hosts which serve the purpose but it has less number of networks as we get with /27;

/26 11111111.11111111.11111111.11000000; i.e 62 hosts and networks 2^26 = this is sure have less # of net then 2^27
/27 11111111.11111111.11111111.11100000; i.e 30 hosts and networks 2^27 =
/28 11111111.11111111.11111111.11110000; i.e 14 hosts which is less then 30

THE QUESTION ASKED ” to allow the maximum number of subnets with as many as 30 hosts each.” From all available answers /27 is the correct options.

Hope this helps.

By: lyra

lyra — Tue, 03 Feb 2015 20:23:22 +0000

Lubna,
.240 = 11110000; 2^4 = 16 subnets 2^4-2 = 14 host; it covers 3 host but max# of net. is 16
.248 = 11111000; 2^5 = 32 subnets 2^3-2 = 6 hosts; it covers 3 host but max# of net is 32

Question specifically asked the maximum number of subnets therefore .248 is correct.

Hope this helps.

By: Anonymous

Anonymous — Sat, 31 Jan 2015 21:06:56 +0000

Q6 today. Praise allah for i have passed

By: gourav

gourav — Thu, 29 Jan 2015 22:17:27 +0000

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By: Lubna Khan

Lubna Khan — Tue, 27 Jan 2015 18:17:10 +0000

Hi SB and Ali

In Q3 , why not option d with 240?

i would go with “D” as this would give u most of the IPs which you cant have from the rest of the options, mask of 255.255.255.240 would mean bit mask of 28 which would mean 4 bits left for hosts that would bring you 2 to the power of 4 minus 2 = 14 hosts. Options a, c and e would get you 2 IPs and b would get u only 6 IPs using the same formula that we opted for d.

whats your opinion?

By: blopslomp

blopslomp — Mon, 26 Jan 2015 16:44:41 +0000

San,

You are given a laminated piece of paper to write on with a marker. You can do whatever you want on that during the test.

By: Stephen

Stephen — Mon, 26 Jan 2015 09:45:50 +0000

Hi San,

As far as I know,, calculators and papers are not allowed

By: san

san — Sat, 24 Jan 2015 19:50:27 +0000

my friends can i use calculator or paper in exam to solve subnetting questions???

By: Junior

Junior — Thu, 22 Jan 2015 05:56:27 +0000

Q9 easy way.
/24 contains 256 addresses.
/23 contains 512 addresses.

Each /30 will consume 4 addresses (1 network, 2 valid addresses, 1 broadcast)

So, 256/4 = 64 (not enough)
512/4 = 128 (enough for 113 subnets)

By: murad

murad — Mon, 19 Jan 2015 19:58:18 +0000

Q8 in my exam today . Passed

By: Guru Gulab

Guru Gulab — Fri, 16 Jan 2015 15:40:18 +0000

thank you …

By: Anonymous

Anonymous — Fri, 16 Jan 2015 12:28:50 +0000

Q9.
Another way to understand this trick question. It says point-to-point links, ok, what do you need to assign ? Ip address, so you need 113 hosts instead of saying subnets. If you have to borrow 0s from the 1s, so you are getting hosts.
2^7=128, then you have to borrow 7 bits from the mask /30. You can do 30-7= 23 or /23. So thats finish.

By: Guru Gulab

Guru Gulab — Thu, 15 Jan 2015 16:48:58 +0000

Can Someone Explain Q9….subnet mask

By: Faizan

Faizan — Sun, 11 Jan 2015 23:08:30 +0000

Can Someone explain Q7 please.

By: SB

SB — Sat, 10 Jan 2015 18:25:35 +0000

For q3, if we calculate no. of usable hosts for 255.255.255.252, it comes out to be 2. We need to provide IP addresses for 3 servers. Thus options with 252 are eliminated. Now out of .248 and .240 , ,248 provides more number of subnets. Hence ‘B’. Please suggest if otherwise.

By: Anonymous

Anonymous — Thu, 08 Jan 2015 06:54:57 +0000

pls ,explain q3, why choose (B).

By: uday

uday — Sun, 28 Dec 2014 12:11:23 +0000

Q5 today

By: for Q10

for Q10 — Sat, 20 Dec 2014 19:58:35 +0000

for commenter below me that ask Q.10,
A is correct, B,C,D,E= wrong
subnet was /28 that means 0,16,32,64,..etc
and ip host : 172.16.28.252 so that between 172.16.16.0-172.16.31.255

By: HM

HM — Fri, 19 Dec 2014 11:05:38 +0000

q3,q7,q8 ….Dec 18

By: Q10

Q10 — Thu, 18 Dec 2014 09:52:00 +0000

Hi all, can someone please explain the question 10, I calculate and I find that network 172.16.0.0 and 172.16.16.0 are true not only 172.16.16.0
thx

By: ali

ali — Tue, 16 Dec 2014 21:25:37 +0000

for through understanding study “binary to decimal” conversion. 3 ips are required from same sub-network. write on paper 128,64,32,16,8,4,2,1. Now write “1″ under each figure till 8. this means in last octet only 11111000.now add 128+64+32+16+8=248, if binary to decimal conversion is done it makes 11111000=248.

By: riyas

riyas — Tue, 16 Dec 2014 07:24:31 +0000

can you explain q3 please brief

By: Anonymous

Anonymous — Wed, 10 Dec 2014 14:41:52 +0000

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By: 9tut

9tut — Tue, 09 Dec 2014 01:06:50 +0000

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By: Enzo

Enzo — Mon, 08 Dec 2014 21:55:57 +0000

Yesterday

By: Enzo

Enzo — Mon, 08 Dec 2014 21:55:38 +0000

Q8 and Q9

By: Zaidam

Zaidam — Thu, 04 Dec 2014 22:39:57 +0000

@anush…
Question 5 stated that the network is 192.168.20.24/29 .
1) /29 means 255.255.255.248
2) This mask tells you that the block is 8 because (256-248=8), so the range of addresses will be between 24 and 32.
3) Go back to the address 192.168.20.24 and apply what you have found . the first usable address is 192.168.20.25 /29 which will be assigned to the router (the gateway)
4) The last usable address in the range is 192.168.20.30 will be assigned to the server.

Hope that will clear why C is the correct answer.

By: anush

anush — Tue, 02 Dec 2014 17:50:22 +0000

isnt for Q5, the answer is d ?
even the subnet is .240 and not 248.
do explain if im wrong, thanks much.

By: anush

anush — Tue, 02 Dec 2014 17:49:19 +0000

isnt for

By: Fatima

Fatima — Tue, 02 Dec 2014 16:54:16 +0000

@ Md. Rezaul Islam…..

They are asking about gateway 192.168.5.33/28 which means the network is 192.168.5.32/28 because it is a multiple of 16. (16 * 2 = 32) . add 32+16 = 48 hosts can be accommodated in this network. So any host lying between 33-47 will be a part of this network. .47 will be a broadcast address?

By: Md.Rezaul Islam

Md.Rezaul Islam — Tue, 02 Dec 2014 03:41:07 +0000

Hi,
I can’t understand it please help me by details explain.
Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)
A. 115.64.8.32
B. 115.64.7.64
C. 115.64.6.255
D. 115.64.3.255
E. 115.64.5.128
F. 115.64.12.128
Answer: B C E

By: Md.Rezaul Islam

Md.Rezaul Islam — Tue, 02 Dec 2014 03:38:23 +0000

Hi,
I can’t understand it please help me by details explain.
The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the
valid IP address of this host?
A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14
Answer: C

By: Anonymous

Anonymous — Tue, 02 Dec 2014 03:37:20 +0000

Hi,
I can’t understand it please help me by details explain.

The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the
valid IP address of this host?
A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14
Answer: C

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By: x

Sat, 29 Nov 2014 17:46:33 +0000

I dont understand Q7, can any explain me please?

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By: Marco

Marco — Mon, 17 Nov 2014 16:11:38 +0000

Agree with Will and Rod: the right answer for Q4 is C, not D. Sure about this!

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Parth Shah — Thu, 13 Nov 2014 10:16:13 +0000

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By: kylie

kylie — Fri, 07 Nov 2014 20:55:24 +0000

can someone explain to me better the question 6? i got the answer, but too different in explanation. i assume it as class B, even the /24 mask. then the network address range is 172.16.32.0-172.16.164.0, its more than 128 subnet so i thought thats why we need /16 mask that has 256 subnet. but what i understand in the explanation is that if theres /18, that would be the better answer. :( pls help me with this. thanks

By: rod

rod — Fri, 07 Nov 2014 07:38:40 +0000

I agree with will. The question was asking for the whole network’s mask and not for each subnet. please see to this guys. thanks

By: Will

Will — Tue, 04 Nov 2014 08:54:20 +0000

Question 4 is asking the size of the mask of a network, which subneted will give 8 subnets able to handle 5 to 26 hosts, not the size of the mask of one resulting subnet! The answer is definitively /24.