Subnetting Questions and Answers

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.

Answer: C

A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.

Answer: A

A /28 is a 255.255.255.240 mask. Let's count to the ninth subnet (we need to find the broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting at 16 (remember, the question stated that we will not use subnet zero, so we start at 16, not 0), 16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128 and the next subnet is 144, so our broadcast address of the 128 subnet is 143. This makes the host range 129-142. 142 is the last valid host.

Answer: D

First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).

Answer: C

A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.

Answer: A

Correct answer: 6

A /29 leaves three host bits, so \[ 2^{3} - 2 = 8 - 2 = 6 \] usable addresses - the block of eight less the subnet address and the broadcast. The router interface is one of the six, so five hosts may join it. The 8 offered is the block size before those two are removed.

Answer: B

The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.