Comments on: CCNA 1 Chapter 6 V4.0 Answers

By: Hicham

Hicham — Sun, 10 Mar 2013 03:56:26 +0000

Greetings Rusmir,

So in regards to question number 3, look at the address and the subnet Mask 192.168.10.0/29, first thing we will do is find the network addresses and their range so we can look at the 5 th subnet the question is talking about, the mask is /29 so it’s 24+5 we borrowd 5 bits to create the subnetworks ( we don’t care how many for the sake of the question ) so we 3 bits left for the host and 2*3=8 so the network addresses ( only curious about the 5 th ) are 192.168.10.0 /192.168.10.8/192.168.10.16/192.168.10.24/192.168.10.32/192.168.10.40
The fifth network address is what interest us, and that’s 192.168.10.32, the first and last host addresses aren’t used ( network and broadcast ) so we are left with this range :
192.168.10.33 to 192.168.10.38 ( .39 is the broadcast address and .32 is the network address ) so now we know the server is getting the last address 192.168.10.138 and the router is getting the first 192.168.10.33 and since we borrowed 5 bits, the subnet Mask is 255.255.255.248

So this is why IP address: 192.168.10.38 subnet mask: 255.255.255.248 default gateway: 192.168.10.33 is the right answer.

Thank you.

By: Rusmir

Rusmir — Wed, 30 Jan 2013 11:31:24 +0000

Can anyone please explain Q3 and Q4?
Thanks?

By: Rusmir

Rusmir — Mon, 28 Jan 2013 09:09:55 +0000

Q4
Can anyone explain why is network prefix 27? I don’t understand?

By: Claudio Sousa

Claudio Sousa — Mon, 24 Sep 2012 18:33:13 +0000

you have to reserve 6 bits if you want use 64 hosts. So, you have 6 for hosts and 2 for network. It means 128 + 64 = 192. Thats why : 255.255.255.192.

By: Justin

Justin — Thu, 20 Sep 2012 02:23:40 +0000

The X got moved and is suppose to be under the 128 Network (TOP LINE) and 2 Host(BOTTOM LINE)

By: Justin

Justin — Thu, 20 Sep 2012 02:20:39 +0000

Number 9, they are asking what network is host 172.25.67.99 /23 on and put it in binary form. So if you have a CIDR of /23 that means that you are in the 3rd octet. On a Class B address that would have you in the 3rd octet with the first network address of 172.25.0.0. So If you follow by your number line of the third octet

2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1
X
So where the X marks the spot of your 7th spot in the 3rd octet you have 128 subnets with 2 hosts in each. So with saying that your going in increments of 2 hosts at a time. 172.25.0.0,172.25.2.0,172.25.4.0 etc. for 128 networks. Which means the IP address 172.25.67.99 which is a broadcast would fall under the 172.25.66.0 Network. So the answer would be

172 25 60 0
10101100. 00011001.01000010.00000000

By: TheMan

TheMan — Wed, 11 Jul 2012 01:35:18 +0000

0/192/224/240/248/252/254/255 and the only usable subnets…

By: amar

amar — Fri, 18 May 2012 11:58:08 +0000

I have enquirer anbot q2.
the maximum hosts are 33, that mean i should reserve 7 bits for hosts (64 host)and 1 bit for network and the answer should be
255.255.255.128

why you selected
255.255.255.192??

Thanks

By: Renegade

Renegade — Mon, 06 Feb 2012 08:05:55 +0000

Is there something wrong with the following:
16. Which three IP addresses are private? (Choose three.)

10.35.66.70

192.168.99.5

172.18.88.90

I was going to check for the answers. It gave me 1 point less and then pointed me to the “public and private addresses” section. I don’t understand… I ‘m pretty sure i had gotten it right. Plus I also checked other websites and apparently I did right. What the hell?

By: susu

susu — Sat, 29 Oct 2011 12:22:55 +0000

good job