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Join Aragoen Celtdra in his pursuit towards CCNA stardom.
Which of the following IP addresses fall into the CIDR block of 110.68.4.0/22? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.68.6.255
D. 110.68.3.254
E. 110.68.5.128
F. 110.68.12.128
Answer: B, C, E.
A Class A network address with a /22 is 255.255.252.0. The subnets in the third octet are 0, 4, 8, 12, etc. The network address in the question is 110.68.4.0, with a broadcast of 110.68.7.255, since the next subnet is 110.68.8.0. Answers B, C, and E are correct host IDs.
The binary equivalent of the prefix notation for /22 is 11111111.11111111.11111100.00000000. Converting that to decimal, you would get 255.255.252.0.
For the network part, just copy the first 2 octets of the subnet number, that is: 110.68
For the host part (the 4th octet, since the binary value is 00000000, the decimal value is: 0
To solve for the 3rd octet, subtract 252 , the 3rd octet of the subnet mask from 256:
256
-252
4
That means that the 3rd octet of the subnet will increment by 4 for each consecutive subnet blocks. Looking at the list of the subnets we can find specific hosts that fall in the given subnet block above. Here are the first few range of subnets:
110.68.0.0 - 110.68.3.255
110.68.4.o - 110.68.7.255
110.68.8.o - 110.68.11.255
110.68.12.0 - 110.68.15.255
and so on...
Going back to the original question, it asks what IP address from the given choices fall in the subnet block of 110.68.4.0. If you look at the list above, the second range is what we want to compare the choices from. Using that, B,C and E are the correct answers.
11.) Which of the following IP addresses fall into the CIDR block of 110.68.4.0/18? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.67.6.255
D. 110.66.3.254
E. 110.65.5.128
F. 110.64.12.128
Answer: B, C, E. A Class A network address with a /18 is 255.255.192.0. The subnets in the third octet are 0, 64, 128, 192. The network address in the question is 110.64.0.0, with a broadcast of110.64.127.255, since the next subnet is 110.64.128.0. Answers B, C, and E are correct host IDs.
Here's how I solved it:
Prefix /18 --> 11111111.11111111.11000000
Decimal equivalent: 255.255.192.0
The zero subnet for the given network should be 110.68.0.0. Based on what we know of Network Classes, this is a class A network with the first octets defining the network. In the explanation given, it mentions that the network address is 110.64.0.0. That doesn't make sense to me because even though it is a Class A network, the first two octets can't change since the first two octets of the given mask is 255.255.
Using Wendell Odom's technique in the ICND1 exam guide:
256 --> Total number of decimal values in an octet
-192 --> decimal value of the masks 3rd octet.
= 64 --> The subnet blocks for the 3rd octet.
Therefore the range of IP addresses for this subnet block are:
110.68.0.0 --> Subnet number
110.68.64.0 --> Next non-zero subnet
Therefore the valid IPs for host are 110.68.0.1 - 110.68.63.254. The only IP addresses in the choices given that fall under this range are choices A and B.
Am I wrong on this?!?!
update (03.11.08):
Link to a few responses from Techexams.net forums:
http://www.techexams.net/forums/viewtopic.php?t=32073
Right now I'm waiting for another 2950 to arrive.
As far as what my plans are on how I'm going to desing my logical topology, it's coming along. This whole thing is a good learning experience for me. As much fun as it is to receive a new box in front of your door every other day, it's equally fun researching what these babies can and can't do; what parts, modules, add-ons to get; what kind of cabling works for what, and all that jazz. Basically it's forcing me to learn without feeling forced.
Up next is a mount rack to mount all these babies to and either a 2509 or 2511 access server.