Comments on: CCNA 4 Practice Final Exam V4.0 Answers

By: Rhezus

Rhezus — Thu, 27 Dec 2012 17:52:01 +0000

You are right. Gateway address must be any of the dhcp excluded addressess and the broadcast is .63 which is also already included in the exclusion list.

By: Zzz

Zzz — Wed, 19 Dec 2012 13:11:25 +0000

Q.31:
I think it should be answer C: ” Automatic configuration from the network will be prevented”.

By: JD

JD — Mon, 17 Sep 2012 08:52:01 +0000

yes it can if you use Extended ACL. It should be Fa0/0, only way to control reaching both interfaces.

By: JD

JD — Mon, 17 Sep 2012 08:50:21 +0000

No, Standard ACL position closet to the Source. Extended are closet to Destination. And the question says the same ACL permits network 192.168.0.0 to reach 192.168.1.0, putting the ACL on Serial interface can not control traffic between those 2 networks. Therefore, putting of Fa0/0 Inbound it can do both, stop traffic from reaching S0/0 and allow traffic to reach Fa0/1.

By: pedroantonio

pedroantonio — Fri, 10 Aug 2012 15:40:49 +0000

But they are on the same network (.5 and .6 are two only hosts ip for that network, .4 is the network adress and .7 is the broadcast).

You can see that the keychain string (wich is case-sensitive) it’s different between the routers. It should be the same; authentication fails because of that, then routing updates are not exchanged.
Therefore the right answer is

“The authentication key strings do not match.”

By: Mat

Mat — Sun, 17 Jun 2012 21:46:51 +0000

internal*

By: Mat

Mat — Sun, 17 Jun 2012 21:45:55 +0000

Q.41:

Should be “The serial interfaces of routers A and B are not on the same network.”

By: Mat

Mat — Sun, 17 Jun 2012 21:44:33 +0000

The server shouldn’t know the internet IP of the router behind NAT…

By: Mat

Mat — Sun, 17 Jun 2012 21:42:17 +0000

If you changed DLSAM, then you’d also know that distance is a factor for speed in the DSL community.

By: Mat

Mat — Sun, 17 Jun 2012 21:39:03 +0000

It’s like you can assign max 63(/26) – 32(start address) – 12(dhcp excluded) – 2(subnet and broadcast) = 17