1. E. This is a Class B network address with 12 bits of subnetting—8 in the third octet and 4 in the fourth octet. The subnet in the third octet is 10, and the subnets in the fourth octet are 256 – 240 = 16, 32, 48, etc. Since the fourth octet is using 22, the host is in the 16 subnet, and since the next subnet is 32, the broadcast address for the 16 subnet is 31. The valid host range is the numbers in between, or 17–30.
2. C. This is a Class B network address with 10 bits of subnetting—8 in the third octet and 2 in the fourth octet. The subnet in the third octet is 8, and the subnets in the fourth octet are 256 – 192 = 64, and 128. However, as long as all the subnet bits in the third octet are not all on at once, the subnets in the fourth octet really can be 0, 64, 128 and 192. This means that the host is in the 128 subnet and since the next subnet is 192, our broadcast address is 8.191.
3. D. This is a Class C network address with 5 bits of subnetting. The valid subnets are 256 – 248 = 8, 16, 24, 32, 40, etc. Since the host ID is 33, we are in the 32 subnet. The next subnet is 40, so our broadcast address is 39.
4. C. Take a look at the answers and see which subnet mask will give you what you need for subnetting. 252 gives you 62 subnets, 248 gives you 30 subnets, 240 gives you 14 subnets, and 255 is invalid. Only the third option (240) gives you what you need.
5. B. If you use the mask 255.255.255.0, that only gives you 8 subnet bits, or 254 subnets. You are going to have to use 1 subnet bit from the fourth octet, or 255.255.255.128. This is 9 subnet bits (29 – 2 = 510).
6. B. To answer this, you must be able to determine which Class C mask provides how many hosts and subnets. The 255.255.255.0 mask provides one network with 254 hosts. The 255.255.255.192 provides two subnets each with 62 hosts. The 255.255.255.224 provides 6 subnets, each with 30 hosts, and the 255.255.255.248 mask provides 30 subnets, each with 6 hosts.
7. B. The Class C mask of 255.255.255.192 provides two subnets (four if you are using subnetzero— which Cisco is not!), each with 62 hosts.
8. D. The third octet is used for all subnets, and the fourth octet is used only for hosts. 8 bits for subnetting, 8 bits for hosts. However, a subnet is already listed, so you have one subnet with 254 hosts. If the question stated 172.16.0.0/24, then the answer would be 254 subnets each with 254 hosts.
9. B. This one takes some thought. 255.255.255.0 would give you 254 hosts each with 254 subnets. Doesn’t work for this question. 255.255.254.0 would provide 126 subnets, each with 510 hosts; the second option looks good. 255.255.252.0 is 62 subnets, each with 1022 hosts. So 255.255.254.0 is the best answer.
10．B. To answer this, you must know that /28 is 255.255.255.240. 256 – 240 = 16. You subtract 2 from this number for all subnet bits and host bits on/off, so the answer is 14 subnets with 14 hosts each.