CCNA DC FAQ: Anayzing Classful IPv4 Networks
Figure: Numbers and Sizes of Class A, B, and C Networks
Q1. Which of the following are not valid Class A network IDs? (Choose two answers.)
a. 1.0.0.0
b. 130.0.0.0
c. 127.0.0.0
d. 9.0.0.0
Answer: B and C. Class A networks have a first octet in the range of 1–126, inclusive, and their network IDs have a 0 in the last three octets. 130.0.0.0 is actually a Class B network (first octet range 128–191, inclusive). All addresses that begin with 127 are reserved, so 127.0.0.0 is not a Class A network.
Q2. Which of the following are not valid Class B network IDs?
a. 130.0.0.0.
b. 191.255.0.0.
c. 128.0.0.0.
d. 150.255.0.0.
e. All are valid Class B network IDs.
Answer: E. Class B networks all begin with values between 128 and 191, inclusive, in their first octets. The network ID has any value in the 128–191 range in the first octet, and any value from 0–255 inclusive in the second octet, with decimal 0s in the final two octets. Two of the answers show a 255 in the second octet, which is acceptable. Two of the answers show a 0 in the second octet, which is also acceptable.
Q3. Which of the following are true about IP address 172.16.99.45’s IP network? (Select two answers.)
a. The network ID is 172.0.0.0.
b. The network is a Class B network.
c. The default mask for the network is 255.255.255.0.
d. The number of host bits in the unsubnetted network is 16.
Answer: B and D. The first octet (172) is in the range of values for Class B addresses (128–191). As a result, the network ID can be formed by copying the first two octets (172.16) and writing 0s for the last two octets (172.16.0.0). The default mask for all Class B networks is 255.255.0.0, and the number of host bits in all unsubnetted Class B networks is 16.
Q4. Which of the following are true about IP address 192.168.6.7’s IP network? (Select two answers.)
a. The network ID is 192.168.6.0.
b. The network is a Class B network.
c. The default mask for the network is 255.255.255.0.
d. The number of host bits in the unsubnetted network is 16.
Answer: A and C. The first octet (192) is in the range of values for Class C addresses (192–223). As a result, the network ID can be formed by copying the first three octets (192.168.6) and writing 0 for the last octet (192.168.6.0). The default mask for all Class C networks is 255.255.255.0, and the number of host bits in all unsubnetted Class C networks is 8.
Q5. Which of the following is a network broadcast address?
a. 10.1.255.255
b. 192.168.255.1
c. 224.1.1.255
d. 172.30.255.255
Answer: D. To find the network broadcast address, first determine the class, and then determine the number of host octets. At that point, convert the host octets to 255 to create the network broadcast address. In this case, 10.1.255.255 is in a Class A network, with the last three octets as host octets, for a network broadcast address of 10.255.255.255. For 192.168.255.1, it is a Class C address, with the last octet as the host part, for a network broadcast address of 192.168.255.255. Address 224.1.1.255 is a Class D address, so it is not in any unicast IP network, so the question does not apply. For 172.30.255.255, it is a Class B address, with the last two octets as host octets, so the network broadcast address is 172.30.255.255.
Q6. Which of the following is a Class A, B, or C network ID?
a. 10.1.0.0
b. 192.168.1.0
c. 127.0.0.0
d. 172.20.0.1
Answer: B. To find the network ID, first determine the class, and then determine the number of host octets. At that point, convert the host octets to 0 to create the network ID. In this case, 10.1.0.0 is in a Class A network, with the last three octets as host octets, for a network ID of 10.0.0.0. For 192.168.1.0, it is a Class C address, with the last octet as the host part, for a network ID of 192.168.1.0. Address 127.0.0.0 looks like a Class A network ID, but it begins with a reserved value (127), so it is not in any Class A, B, or C network. 172.20.0.1 is a Class B address, with the last two octets as host octets, so the network ID is 172.20.0.0.
Related
