Subnetting Flashcards
7 cards from real ICND1 practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Subnetting flashcards as text
Which address is the network address of the subnet containing host 192.168.100.77/28?
Answer: 192.168.100.64
A /28 has a block size of 16; 77 ÷ 16 = 4 remainder 13, so the subnet starts at 4 × 16 = 64.
What CIDR notation represents the subnet mask 255.255.255.192?
Answer: /26
255.255.255.192 has 26 bits set (24 + 2 from the last octet: 11000000), giving /26.
You have 10.0.0.0/8 and need to create subnets for WAN links using the least address space. Which mask is most efficient for a two-router serial link?
Answer: /30
A /30 provides exactly 2 usable host addresses, which is precisely what a point-to-point WAN link requires.
If a router has the interface IP 172.20.0.1/23, which network does it belong to?
Answer: 172.20.0.0
A /23 has a block size of 2 in the third octet; 0 ÷ 2 = 0, so the network is 172.20.0.0.
How many subnets and hosts per subnet does 10.0.0.0/8 subnetted to /12 provide?
Answer: 16 subnets, 1,048,574 hosts each
12 - 8 = 4 borrowed bits → 2^4 = 16 subnets; 32 - 12 = 20 host bits → 2^20 - 2 = 1,048,574 hosts each.
A technician enters 'ip address 192.168.1.33 255.255.255.224' on an interface. What is the valid host range for this subnet?
Answer: 192.168.1.33 – 192.168.1.62
255.255.255.224 is /27 with a block of 32; the subnet containing .33 starts at .32, so hosts are .33–.62 (.63 is broadcast).
Which of the following is TRUE about the subnet 0.0.0.0/0?
Answer: It represents the default route matching all IP addresses
0.0.0.0/0 has no bits fixed, so it matches every possible IP address and is used as the default route.