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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.