If there is one absolute truth about the Cisco CCNA exam, it is this: You must be able to subnet quickly and accurately. You will not have time to draw out massive binary conversion tables during the test. Cisco deliberately words their questions to trick candidates who haven't mastered the fundamentals.
Below are the five most common styles of subnetting questions you will face on the exam, how to decode what they are actually asking, and how to solve them fast.
Practice Makes Perfect
Generate your own practice questions by typing random IP addresses into our tool, guessing the boundaries, and then verifying your answers.
Use the All-in-One Calculator →Question 1: "What is the valid host range for the subnet containing the IP address 192.168.4.67/26?"
How to solve it: A
The IP
/26 mask leaves 6 host bits. The block size (magic number) is 2^6 = 64. The subnets increment by 64: 0, 64, 128, 192. The IP
.67 falls between .64 and .128. Therefore, the network address is 192.168.4.64. The broadcast is one less than the next subnet, so .127. The valid hosts are between those two numbers: 192.168.4.65 through 192.168.4.126.
Question 2: "Which of the following IP addresses can be assigned to a host on the 10.1.1.0/28 subnet?"
How to solve it: A
Any option they provide that is
/28 mask means the block size is 16 (from 2^4). The network starts at 10.1.1.0. The next network is 10.1.1.16. The broadcast for this subnet is 10.1.1.15. Any option they provide that is
.0 or .15 is invalid. You must select an IP from 10.1.1.1 to 10.1.1.14.
Question 3: "You need to subnet a Class B network to provide at least 100 subnets, each supporting at least 500 hosts. Which mask should you use?"
How to solve it: A Class B default is
Subtract the 9 host bits from the 32 total bits: 32 - 9 = 23. You need a /23 subnet mask (which is 255.255.254.0).
/16. You need 500 hosts. The formula 2^h - 2 must be ≥ 500. 2^9 = 512. So you need 9 host bits. Subtract the 9 host bits from the 32 total bits: 32 - 9 = 23. You need a /23 subnet mask (which is 255.255.254.0).
Question 4: "Router A cannot ping Router B. Router A's IP is 172.16.1.129/25 and Router B's IP is 172.16.1.126/25. Why?"
How to solve it: This is a classic boundary trick. A
Router A (.129) is on the second subnet. Router B (.126) is on the first subnet. They are not on the same network and therefore cannot ping each other without a default gateway route between them.
/25 has a block size of 128. The first subnet is .0 to .127. The second subnet is .128 to .255.Router A (.129) is on the second subnet. Router B (.126) is on the first subnet. They are not on the same network and therefore cannot ping each other without a default gateway route between them.
Question 5: "What is the broadcast address of the subnet for host 10.255.255.33 /27?"
How to solve it: A
The broadcast is exactly one number less than the next subnet. Therefore, the broadcast is 10.255.255.63.
/27 has a block size of 32. The subnets are 0, 32, 64. The IP .33 falls in the .32 subnet. The next subnet starts at .64. The broadcast is exactly one number less than the next subnet. Therefore, the broadcast is 10.255.255.63.