Subnetting Questions and Answers

Answer: A

In classful addressing the class is decided by the first octet. The range 1 to 126 is class A, 128 to 191 class B, 192 to 223 class C and 224 to 239 class D. The address 123.23.156.4 has first octet 123, which lies in 1 to 126, so it belongs to class A. It is not class B, which would need a first octet of 128 or more, nor class C or D. The keyed answer is A.

Answer: C

The address 200.200.200.200 belongs to Class C. In IPv4 the class is decided by the first octet: 0 to 127 is Class A, 128 to 191 is Class B, 192 to 223 is Class C, and 224 to 239 is Class D (multicast). The first octet here is 200, which lies in the Class C range.

It is not Class A or B because 200 is above 191, and not Class D because 200 is below 224.

Answer: B

The address 190.255.254.254 belongs to Class B, option (B). In classful addressing the class is read from the first octet: 0 to 127 is Class A, 128 to 191 is Class B, 192 to 223 is Class C, 224 to 239 is Class D (multicast) and 240 to 255 is Class E. The first octet here is 190, which falls in 128 to 191, and in binary it begins with the bits 10. Class A would need the first octet below 128, Class C at least 192 and Class D at least 224, so (A), (C) and (D) are wrong.

Answer: A

A Class A address has 7 network bits and 24 host bits. The first bit of the first octet is 0, which leaves 7 bits for the network number in the first octet, and the other three octets, 24 bits, are for the host.

Fourteen network bits and 16 host bits is the split of Class B, which starts with the bits 10. Fifteen and sixteen, and sixteen and sixteen, are not splits of any address class, so the other options are wrong.

Answer: B

The subnet address is 201.14.78.64. It is obtained by a bitwise AND of the address and the mask. Only the last octet is affected, since the first three octets of the mask are 255.

\[ 65 = 01000001_{2} \]

\[ 224 = 11100000_{2} \]

\[ 01000001 \wedge 11100000 = 01000000 = 64 \]

So the subnet address is 201.14.78.64. The value 201.14.78.32 would need a last octet in the range 32 to 63, and .65 and .224 are the host address and the mask, which are not the network address.

Answer: C

The network can handle 4096 hosts.

The mask 255.255.240.0 has \(8 + 8 + 4 = 20\) network bits, so 12 bits remain for hosts and \(2^{12} = 4096\) addresses are available.

1024, 2048 and 8192 correspond to 10, 11 and 13 host bits. The count of usable hosts is often given as 4094 because the all-zero and all-ones host addresses are reserved, but 4096 is the number of addresses in the block and is the only option near that value.

Answer: B

Correct answer: 8 subnets, 8,190 hosts each

172.16.0.0 is a class B network with a natural /16, so a /19 borrows three bits and leaves thirteen: \[ 2^{3} = 8\ \mathrm{subnets}, \qquad 2^{13}-2 = 8190\ \mathrm{hosts}. \] All eight subnets are usable on any modern IOS, and the answers offering seven subnets come from the obsolete rule that excluded the all-zeros subnet.

Answer: C

Correct answer: 127.0.0.1

127.0.0.1 is the loopback address, and pinging it exercises the local TCP/IP stack without touching any network hardware - a reply proves the software is installed and running. 127.0.0.0 is the network address of the loopback block and 127.0.0.255 a broadcast within it; neither is the host address to test.

Answer: A

A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.

Answer: C

A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.