Comments on: New CCNA – Access list Questions

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By: CCNA-seeker

CCNA-seeker — Fri, 08 May 2015 17:26:28 +0000

passed today. Thanks a lot to 9tut. eigrp trouble shooting lab and both ACL labs came.

By: A.k

A.k — Mon, 04 May 2015 11:57:41 +0000

Hello guys, those who said they passed the exam, did you guys pay the 9$ and reviewed from there or did you just studied the questions here for free?

By: gidz

gidz — Mon, 04 May 2015 08:47:52 +0000

but 192.168.146.0 and 192.168.147.0 was stated as a network, ryt?

By: metacortex

metacortex — Mon, 04 May 2015 02:21:24 +0000

@gidz – don’t summarize. Because of the increment when you have a wildcard mask of 0.0.1.255 you’re including 146-147 and 148-149 in the ACL.
For example:
192.168.146.0/23 (subnet mask 255.255.254.0 / wildcard mask 0.0.1.255)
192.168.146.1 – first host
192.168.147.254 – last host
192.168.147.255 – broadcast address

By: gidz

gidz — Sat, 02 May 2015 11:32:17 +0000

I cant understand question 2
If i try to summarize the networks 146-149, i get the block size of 1, i cant figure out why are you guys trying to summarize network 192.168.146.0 and 192.168.147.0 together and separate them from 192.168.149.0 and 192.168.149.0?…

By: gidz

gidz — Sat, 02 May 2015 11:15:32 +0000

192.168.147.0 can be a network with a block size of 1 means with /24 prefix…

By: SaedAdi

SaedAdi — Fri, 17 Apr 2015 12:38:14 +0000

Q1,2,3,5

By: Iyke

Iyke — Wed, 15 Apr 2015 14:18:54 +0000

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By: sumit7843

sumit7843 — Wed, 15 Apr 2015 07:22:18 +0000

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By: justin13

justin13 — Tue, 07 Apr 2015 22:53:53 +0000

thank you Agya. Its possible I am too literal, but I still think the question is worded improperly

In your explanation you say any IP address can stand for a Host or network address, ‘depending on the mask applied’
in this case, the 255.255.254.0 mask was applied, so technically 192.168.147.0 is not a network in this subnet mask, no?

By: Agya

Agya — Sat, 04 Apr 2015 15:38:58 +0000

Justin13, do not get confused by the phrase “hosts on network 192.168.147.0″. Any IP address can stand for a Host or a Network address, depending on the mask applied. An address with a mask ff.ff.ff.ff can be considered as a Host only. 192.168.147.0 ff.ff.ff.0 is a Network address, whileas 192.168.147.0 ff.ff.fe.0 is a Host address.

By: nesrin

nesrin — Sat, 28 Mar 2015 21:42:43 +0000

In Q3 the answer is no ip access-group 102 out I think they mean there is no such an Acl Active on the interface look at the lesson explanation by clicking on the link at the beginning of the page and you will understand.

By: Siddheshwar

Siddheshwar — Sat, 28 Mar 2015 05:10:45 +0000

please tell me the exam pattern as below:-
1.Number of questions
2.Time of exam in hours
3.marks
4.if there is questions set

By: steve

steve — Thu, 26 Mar 2015 19:29:24 +0000

hey 213

you are leaving out .3 and .4 hosts … with that network statement..

By: steve

steve — Thu, 26 Mar 2015 19:28:30 +0000

Hey 213 look at what you put as the network ip.. that is what is wrong

your using the wrong network ip

3 private IP’s for the LAN Host
Host A 192.168.25.3
Host B 192.168.25.4
******************Host C 192.168.25.5
2 Public IP addresses
198.18.188.25
198.18.188.26
2 Servers were on
172.16.25.5
172.16.25.4
Q1. Write an Access list that will allow host C to access the Finance accounting server via HTTP.
Q2.Other LAN hosts should not access the Finance accounting server but can access the Public web server
Q3. Hosts from the core network should not also access the Finance accounting server but can access the rest.
I tried creating an ACL with the commands below but always got an error:
under config mode:
#access-list 1 permit tcp 192.168.25.5 0.0.0.3 eq 80
#access-list 1 permit 192.168.25.5 0.0.0.3
#access-list 1 permit 192.168.25.5 0.0.0.3 eq 80
#access-list 1 permit tcp 192.168.25.5 0.0.0.3 eq80
#access-list 1 permit 192.168.25.5 0.0.0.3 eq80
Got error on the above.

By: Peter

Peter — Thu, 26 Mar 2015 15:00:15 +0000

Q1,Q5

By: justin13

justin13 — Wed, 25 Mar 2015 17:50:41 +0000

hello 9tut. question:

the only deceiving part of question #2 is the following:
/23= 255.255.254.0
128 subnets/ block size=2
192.168.146.0 network covers from 192.168.146.1-192.168.147.254
192.168.148.0 network covers from 192.168.148.1-192.168.149.254

Technically, 192.168.147.0 is not a network because it resides in the network 192.168.146.0
I was able to figure the answer out based solely on the wildcard mask, and process of elimination
Am I missing something because the question phrases it as “hosts on network 192.168.147.0″
this is a bit deceiving

By: rusy

rusy — Sat, 21 Mar 2015 15:56:02 +0000

How many times can u retake ccna afta u fail???

By: Apush

Apush — Wed, 18 Mar 2015 12:03:54 +0000

Passed my CCNA exam today (18th Mar)… Q2, Q4 and Q7 in exam

By: subodh

subodh — Thu, 12 Mar 2015 03:56:10 +0000

mostly which chapter???

By: subodh

subodh — Thu, 12 Mar 2015 03:55:31 +0000

can anyone tel me that in CCNA exam questions of which chapter apperes???

By: Ghada

Ghada — Tue, 24 Feb 2015 08:34:24 +0000

Passed yesterday with score 1000 Thank God

The exam was 51 questions, only one drag and drop about cable types, all questions are very easy just similar to 9tut.

The lab simulations were Eigrp , ACL1 (similar to 9tut exactly) and ACL2 (modification 3 but with host B to access the finance server)

Good luck for everyone, just be relaxed its very very easy :)

questions3,7 were in it

By: Iliyas shaikh

Iliyas shaikh — Mon, 23 Feb 2015 14:10:06 +0000

Thanks to 9 tut I passed today with 972/1000 all question from 9tut

By: Purdue_Alum

Purdue_Alum — Fri, 20 Feb 2015 20:39:36 +0000

Q2 explanation:

Convert IP to binary:
1. 192.168.146.0, 192.168.147.0
192.168.146.0 — 1001 0010
192.168.147.0 — 1001 0011
==> summary address: 1001 001X = 192.168.146.0 /23 (first 2 octets combined are
16 bits, the 3rd octet has 7 bits for total of 23 bits or /23)
Wildcard for /23 = 0.0.1.255 hence the first ACL covers 192.168.146.0 0.0.1.255

2. 192.168.148.0, 192.168.149.0
192.168.148.0 — 1001 0100
192.168.149.0 — 1001 0101
==> summary address: 1001 010X = 192.168.148.0 /23 (first 2 octets combined are
16 bits, the 3rd octet has 7 bits for total of 23 bits or /23)
Wildcard for /23 = 0.0.1.255 hence the second ACL covers 192.168.148.0 0.0.1.255

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jude — Fri, 20 Feb 2015 13:05:56 +0000

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By: javb

javb — Thu, 19 Feb 2015 16:19:18 +0000

Pass today 890 Thanks….

By: Chains

Chains — Mon, 16 Feb 2015 17:36:57 +0000

@noor, only sumary on two networks
192.168.146.0, 192.168.147.0 ….. sumary 1 acl 1
192.168.148.0, 192.168.149.0….. sumary 2 acl 2

By: noor

noor — Mon, 16 Feb 2015 12:16:26 +0000

Hello!
Can any one explain to me Q2?
thanks a lot

By: Che

Che — Sun, 15 Feb 2015 21:05:56 +0000

Q3, Q7 on 15th Feb.

By: Peter

Peter — Fri, 13 Feb 2015 11:48:22 +0000

q1,2,4,6,7 yesterday. passed!

By: Chains

Chains — Wed, 11 Feb 2015 17:11:24 +0000

What’s ACL1Y2 mode 3 ? Pleaseeeeee

By: Anonymous

Anonymous — Thu, 05 Feb 2015 15:42:14 +0000

“permit 172.21.1.129 0.0.0.0″ allow first host
“permit 172.21.1.142 0.0.0.0″ allow last host
“deny 172.21.1.128 0.0.0.15″ prevent all hosts
“permit any” allow all other traffic
You have to put permit any any at the end access-list because default end access-list is deny any any

Also correct answer would be DCBA

By: Mahmoud

Mahmoud — Wed, 04 Feb 2015 03:10:22 +0000

Q5: i think question is wrong how access ends with permit any and he wants to prevent all hosts accept those two ips as he describes??? any one can expline for me please.

By: Anonymous

Anonymous — Sat, 31 Jan 2015 21:04:17 +0000

Q3 & 7 today. Praise allah for i have passed

By: Stephen

Stephen — Thu, 22 Jan 2015 11:03:53 +0000

Hi Ali,

access-list 10 permit ip 192.168.146.0 0.0.1.255>>> means that it allow ip from .146 to .147 since it used the 0.0.1.255, the .1 on 3rd octet of the wildcard mask, 255-1 = 254 (meaning increment of 2), and on the 4th octet 255 means any,so it allow network 192.168.146.0 and 192.168.147.0. I hope you understand it. It is really hard for me to explain -_-

By: Ali

Ali — Thu, 22 Jan 2015 02:57:28 +0000

Hi,
Can someone explain the Q.2 briefly please?

By: murad

murad — Mon, 19 Jan 2015 19:52:48 +0000

Q6 in my exam today . Passed

By: abdel

abdel — Mon, 19 Jan 2015 13:20:55 +0000

Q3 & 7 in my exam today. Passed

By: Noah

Noah — Sun, 18 Jan 2015 20:22:33 +0000

Answer B is correct. This for routed protocols such as IPv4, IPv6 or old IPX

By: Claudio

Claudio — Mon, 05 Jan 2015 09:53:08 +0000

Answer B is correct because in the old days we used to set IPX standard and extended access control list among other legacy layer 3 protocols as Appletalk and DECnet. ;-)

By: Jero

Jero — Sat, 03 Jan 2015 15:25:17 +0000

dear all
regarding question 6 which is
Which statement about access lists that are applied to an interface is true?

A. you can apply only one access list on any interface
B. you can configure one access list, per direction, per layer 3 protocol
C. you can place as many access lists as you want on any interface
D. you can configure one access list, per direction, per layer 2 protocol

Answer: B

the answer for me is weird because the “per layer 3 protocol”, I reviewed the Extended ACL used protocol keywords like igmp, tcp, udp, ip, eigrp, icmp but UDP and TCP are not layer 3 protocols, so how this answer is right

By: uday

uday — Sun, 28 Dec 2014 12:03:14 +0000

Q1,4 today

By: @mcp3228

@mcp3228 — Mon, 22 Dec 2014 17:48:22 +0000

you need to summarize 192.168.146.0 along with 192.168.147.0 together which would give the subnet mask 255.255.254.0 and the wild card mask would be 0.0.1.255.Same would be done separately with the other two networks (48 and 49) hope this have been informative.
Shoot me an email if it’s not comprehensible mahmouddiser@hotmail.com

By: faty

faty — Fri, 19 Dec 2014 11:24:31 +0000

q6 on 11th dec

By: HM

HM — Fri, 19 Dec 2014 11:00:04 +0000

q2,q7…Dec 18

By: mcp3228

mcp3228 — Mon, 15 Dec 2014 22:03:19 +0000

@ 9tut

Im confuse about question #2

I tought it would have been : access-list 10 permit ip 192.168.146.0 0.0.0.255

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By: Lionleap

Lionleap — Tue, 09 Dec 2014 19:48:43 +0000

How do you configure a port security server I’ve been trying but I cant figure it out

By: Enzo

Enzo — Mon, 08 Dec 2014 21:49:13 +0000

Q1, Q5 and Q6 yesterday

By: Dividend

Dividend — Thu, 04 Dec 2014 22:55:06 +0000

@213

Another mistake I have noticed is you may not have been in global config mode. You showed your commands as “#access-list….” when it should have been “(config)#access-list…”

Although you were mistakenly configuring a standard access list instead of an extended, the command “access-list 1 permit 192.168.25.5 0.0.0.3″ would have been accepted as a valid command…unless Cisco was helping you by warning you not to use a standard access list…but that’s is highly unlikely :)

By: SPS

SPS — Wed, 03 Dec 2014 05:43:44 +0000

The correct answer should have been
access-list 100 permit tcp host 192.168.25.5 host 172.16.25.5 eq 80
access-list 100 deny ip any host 172.16.25.5
access-list 100 permit ip any any

int fa X/X
ip access-group 100 out

copy run start

By: 213

213 — Fri, 28 Nov 2014 20:45:25 +0000

Thanks Max, does it also mean that standard access list if that is what was asked it could have taken the command?…

By: Max

Max — Thu, 27 Nov 2014 19:30:44 +0000

@123
You have wrote the ACL the right form, but you have to use numbers above “99″. Number “1″ is to form standard ACL, and will not work anyway.

By: Twista

Twista — Tue, 25 Nov 2014 14:24:31 +0000

I understand that Standard access list is 1-99, 1300-1999 while Extended IP lists 100-199, 2000-2699 does it mean when using a Standard access list you cannot use eq (equate)
and is only on the Extended list? thus this is what was required?

By: Syed Saqib Hasan

Syed Saqib Hasan — Mon, 24 Nov 2014 18:21:45 +0000

how can you use access-list 1 on extended access list..

By: 213

213 — Mon, 24 Nov 2014 16:28:09 +0000

9tut anything???

By: 213

213 — Mon, 24 Nov 2014 15:37:19 +0000

Which topics are under IP services?

By: 213

213 — Mon, 24 Nov 2014 15:36:09 +0000

May they used question 5. and modified it in to a Testlet with other additions

By: 213

213 — Mon, 24 Nov 2014 15:32:38 +0000

Tried
#access-list 1 permit ip 192.168.25.5 0.0.0.3
#access-list 1 permit ip 192.168.25.5 0.0.0.3 eq 80
refused

By: 213

213 — Mon, 24 Nov 2014 15:27:25 +0000

Did the xam today failed most of the questions are on 9tut but not all had some or a few not popular from other topics I’ll share respectively; but good work 9tut thumbs up. (Getting this simulation would have helped me pass). 9tut i had a simulation /Testlet on Access list that requested me to add only three ACL statements on the router whereby there is a Finance accounting server and a Public Web server both with different IP addresses. There’s a host C on a LAN with three other machines. And there’s the Core network with users but users are not indicated only two public IPs give for the core network (No diagram). The Question was a Testlet so you just save on the router and move to the next question. (I had 2 of this got the EIGRP one i remembered abt the issue of passive-interface enabled it and could ping all branches)

Back to the Question

(I can’t remember the IP’s but first octets of the IP’s were correct as below)

3 private IP’s for the LAN Host
Host A 192.168.25.3
Host B 192.168.25.4
Host C 192.168.25.5

2 Public IP addresses
198.18.188.25
198.18.188.26

2 Servers were on
172.16.25.5
172.16.25.4

Q1. Write an Access list that will allow host C to access the Finance accounting server via HTTP.

Q2.Other LAN hosts should not access the Finance accounting server but can access the Public web server

Q3. Hosts from the core network should not also access the Finance accounting server but can access the rest.

I tried creating an ACL with the commands below but always got an error:
under config mode:
#access-list 1 permit tcp 192.168.25.5 0.0.0.3 eq 80
#access-list 1 permit 192.168.25.5 0.0.0.3
#access-list 1 permit 192.168.25.5 0.0.0.3 eq 80
#access-list 1 permit tcp 192.168.25.5 0.0.0.3 eq80
#access-list 1 permit 192.168.25.5 0.0.0.3 eq80
Got error on the above.
This gave me the error incomplete command #access-list 1 permit 192.168.25.5 0.0.0.3
(i can’t remember any subnets give but on may hv been given whereby i took 255.255.255.255-255.255.255.252 = 0.0.0.3 and got 3 or so but doubt it was a /30 coz it only has 2 available IPs but was something like that)

other command i tried
#access-list deny 1 192.168.25 0.0.0.3 rejected the command
#access-list permit

I noted that there were not IPs configured on the router and figured that i wasn’t creating the ACL in not interface. Questions asked were the 3 there was no configure IPs on the interfaces now am figuring maybe they wanted me to configure the IPs on the interface i think there 2 interfaces. I think this question had something wrong on it Can you help
??? I was not able to go through and it was almost 10 or after 5 sadly skipped as i couldn’t penetrate it time really run out :(

By: spidyfox

spidyfox — Sun, 23 Nov 2014 19:11:15 +0000

Q5 on exam 19 th nov

By: Erika

Erika — Sat, 22 Nov 2014 05:24:38 +0000

Q4, Q7 on exam yesterday

By: Mario

Mario — Tue, 18 Nov 2014 20:50:16 +0000

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Thanks,

By: naqi hasan

naqi hasan — Mon, 17 Nov 2014 17:23:33 +0000

hello every one

By: Parth Shah

Parth Shah — Thu, 13 Nov 2014 10:06:23 +0000

Q 4 and 6 on 10th Nov

By: onepiece

onepiece — Wed, 12 Nov 2014 06:34:10 +0000

@sup – the question is to prevent all hosts (except those whose addresses are the first and last IP of subnet 172.21.1.128/28) from accessing the network. it says the first ip and last IP of the subnet ex exempted in blocking. so the arrangement of the command is like this.

C. permit 172.21.1.129 0.0.0.0 – first ip (exempted in blocking)
D. permit 172.21.1.142 0.0.0.0 – last ip (exempted in blocking)
B. deny 172.21.1.128 0.0.0.15 – subnet
A. permit any

By: onepiece

onepiece — Wed, 12 Nov 2014 06:17:47 +0000

Q 3, why the answer is D? access group is pertaining to ip protocol right? anyone could explain me please? thank you

By: aria

aria — Mon, 10 Nov 2014 20:28:31 +0000

q1 today

By: ajikemi35

ajikemi35 — Mon, 10 Nov 2014 03:06:47 +0000

Sup,

I think the answer could be DCBA or CDBA, however no choice with DCBA, so CDBA would be the obvious choice.

By: sup

sup — Sun, 09 Nov 2014 02:51:37 +0000

how Q5 answer is D? plz tell me anyone..

By: darsh

darsh — Fri, 07 Nov 2014 18:16:57 +0000

jesus plz tell us about your exam . mine will be next sunday

By: kam

kam — Thu, 06 Nov 2014 20:54:05 +0000

q-1,3,6 on exam today

By: Jesus

Jesus — Tue, 04 Nov 2014 21:33:15 +0000

My examen is this Friday….Im so nervious

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