网管员子网划分专项练习（附答案）
子网划分专项练习
1. What valid host range is the IP address 172.16.10.22 255.255.255.240 a part of?
A. 172.16.10.20 through 172.16.10.22
B. 172.16.10.1 through 172.16.10.255
C. 172.16.10.16 through 172.16.10.23
D. 172.16.10.17 through 172.16.10.31
E. 172.16.10.17 through 172.16.10.30
2. What is the broadcast address of the subnet address 172.16.8.159 255.255.255.192?
A. 172.16.255.255
B. 172.16.8.127
C. 172.16.8.191
D. 172.16.8.255
3. What is the broadcast address of the subnet address 192.168.10.33 255.255.255.248?
A. 192.168.10.40
B. 192.168.10.255
C. 192.168.255.255
D. 192.168.10.39
4. If you wanted to have 12 subnets with a Class C network ID, which subnet mask would you use?
A. 255.255.255.252
B. 255.255.255.248
C. 255.255.255.240
D. 255.255.255.255
5. If you need to have a Class B network address subnetted into exactly 510 subnets, what subnet
mask would you assign?
A. 255.255.255.252
B. 255.255.255.128
C. 255.255.255.0
D. 255.255.255.192
6. If you are using a Class C network ID with two subnets and need 31 hosts per network, which
of the following masks should you use?
A. 255.255.255.0
B. 255.255.255.192
C. 255.255.255.224
D. 255.255.255.248
7. How many subnets and hosts can you get from the network 192.168.254.0/26?
A. 4 networks with 64 hosts
B. 2 networks with 62 hosts
C. 254 networks with 254 hosts
D. 1 network with 254 hosts
8. You have the network 172.16.10.0/24. How many subnets and hosts are available?
A. 1 subnet with 10 hosts
B. 1 subnet with 254 hosts
C. 192 subnets with 10 hosts
D. 254 subnets with 254 hosts
9. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100 subnets
with about 500 hosts each?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.0.0
11You have a Class C 192.168.10.0/28 network. How many usable subnets and hosts do you have?
A. 16 subnets, 16 hosts
B. 14 subnets, 14 hosts
C. 30 subnets, 6 hosts
D. 62 subnets, 2 hosts
13. You have the network 192.168.10.0/24. How many subnets and hosts are available?
A. 1 subnet with 10 hosts
B. 1 subnet with 254 hosts
C. 192 subnets with 10 hosts
D. 254 subnets with 254 hosts
14. You have a 255.255.255.240 mask. Which two of the following are valid host IDs?
A. 192.168.10.210
B. 192.168.10.32
C. 192.168.10.94
D. 192.168.10.112
E. 192.168.10.127
15. You have a Class B network ID and need about 450 IP addresses per subnet. What is the best
mask for this network?
A. 255.255.240.0
B. 255.255.248.0
C. 255.255.254.0
D. 255.255.255.0
17. Which of the following is a valid host for network 192.168.10.32/28?
A. 192.168.10.39
B. 192.168.10.47
C. 192.168.10.14
D. 192.168.10.54
19. What is the valid host range that host 192.168.10.22/30 is a part of?
A. 192.168.10.0
B. 192.168.10.16
C. 192.168.10.20
D. 192.168.0.0
Answers to Review Questions
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. B. 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.
11．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.
13. B. This is a Class C network using a default mask. This provides a simple single network with
254 hosts.
14. A, C. To answer this, just do 256 – 240 = 16. Keep adding 16 together until you reach the subnet
mask value. 16 + 16 = 32. We’ll keep adding 16 until we have all our subnets: 16, 32, 48, 64, 80,
96, 112, 128, 144, 160, 176, 192, 208, 224. The broadcast addresses for each subnet are the
numbers right before the next subnet and are not valid hosts. The answers are 210 and 94.
15. C. Start with 255.255.255.0. This provides 254 subnets, each with 254 hosts. Move the subnet
bits right if you need more subnets; move them left if you need more hosts. Since we need more
hosts, we’re going to take away subnet bits. The next mask then is 255.255.254.0, which provides
7 subnet bits and 9 host bits, or 126 subnets, each with 510 hosts.
17. A. You have to know that /28 is 255.255.255.240. 256 – 240 = 16, 32, 48, etc. The host ID of
34 is in the 32 subnet; the next subnet is 48, so the broadcast address is 47. The valid host range
is 33–46, so answer A is correct.
19. C. This is as easy as they get. A /30 is a 255.255.255.252 mask. 256 – 252 = 4, 8, 12, 16, 20, 24. This
host is in the 20 subnet, the broadcast address is 23 and the valid host range is 21 and 22.
