ch03 IP Addressing.pdf - The Cisco Learning Network

46 Chapter 3: IP Addressing
Planning IP Addressing
10. B. Class C networks have 24 bits--this example steals on bit. 2 raised to the power
of 1 equals 2 subnets.
A, C, and D are incorrect.
11. B. There are 3 host bits, with 2 raised to the power of 3 resulting in 8 addresses in a
network, but you lose 2 for the network and directed broadcast address, resulting in 6 host
addresses. You could also subtract 248 from 256, resulting in a total of 8 addresses per network,
of which the first and last are reserved.
A, C, and D are incorrect.
12. B. There are 6 host bits, with 2 raised to the power of 6 resulting in 64 addresses in a
network, but you lose 2 for the network and directed broadcast address, resulting in 62 host
addresses. You could also subtract 192 from 256, resulting in a total of 64 addresses in a
network…but you can’t use the first and the last.
A, C, and D are incorrect.
Figure Out IP Address Components
13. C. There is 1 subnet bit for this Class C network, resulting in two networks: 192.168.37.0
and 192.168.37.128, making 192.168.37.192 a host address. Host addresses for this subnet
range from 192.168.37.129-192.168.37.254.
A is true for 192.168.37.0 and 192.168.37.128. B is true for 192.168.37.127 and
192.168.37.255.
14. C. Network addresses are incrementing by 2 in the third octet. Where 172.17.16.255
is a host address. Host addresses range from 172.17.16.1-172.17.17.254.
A is true for 172.17.16.0. B is true for 172.17.17.255.
15. A. Network addresses are incrementing by 4 in the third octet. Where 10.0.8.0 is a
network address. Other network addresses include 10.0.0.0, 10.0.4.0, 10.0.8.0, 10.0.12.0,
and so on and so forth.
B is true for 10.0.3.255, 10.0.7.255, 10.0.11.255, 10.0.15.255, and so on and so forth.
C is true for 10.0.8.1-10.0.11.255, as well as other host addresses in other subnets.
16. C. Remember that the OUI portion is the first six hexadecimal characters in a MAC
address (Chapter 2). Hex characters are represented in 4 bits in binary. 0=0000, 1=0001,
A=1010, 2=0010, 0=0000, and 4=0100. When you concatenate these together, the result
is 00000001.01000010.0100 (the first six hexadecimal digits).
A converts to 0B for the first byte, making it incorrect. B converts to 00 for the first byte,
making it incorrect. D’s first byte is correct, but the second byte converts to A1 and third byte
to 02, making it incorrect