ICND1 Subnetting 2 — Questions and Answers
Question 1: How many usable host addresses are available in a /27 subnet?
- 30 (Correct answer)
- 32
- 14
- 62
Correct answer: 30
A /27 subnet has 32 total addresses (2^5), minus 2 for network and broadcast, leaving 30 usable hosts.
Question 2: What is the broadcast address for the network 192.168.10.64/26?
- 192.168.10.127 (Correct answer)
- 192.168.10.126
- 192.168.10.128
- 192.168.10.255
Correct answer: 192.168.10.127
A /26 block is 64 addresses; starting at .64, the broadcast is .64 + 63 = .127.
Question 3: Which subnet mask corresponds to the CIDR notation /20?
- 255.255.240.0 (Correct answer)
- 255.255.248.0
- 255.255.252.0
- 255.255.224.0
Correct answer: 255.255.240.0
/20 means 20 bits are set, so the third octet has the top 4 bits set: 11110000 = 240, giving 255.255.240.0.
Question 4: A network administrator needs at least 50 subnets from 172.16.0.0/16. What is the minimum number of bits to borrow from the host portion?
- 6 (Correct answer)
- 5
- 7
- 4
Correct answer: 6
2^6 = 64 subnets, which satisfies the requirement of at least 50; 2^5 = 32 is insufficient.
Question 5: What is the network address for the host IP 10.4.77.130/25?
- 10.4.77.128 (Correct answer)
- 10.4.77.0
- 10.4.77.64
- 10.4.77.192
Correct answer: 10.4.77.128
A /25 mask is 255.255.255.128; ANDing 130 with 128 gives 128, so the network is 10.4.77.128.
Question 6: Which of the following IP addresses belongs to the subnet 192.168.5.128/26?
- 192.168.5.190 (Correct answer)
- 192.168.5.64
- 192.168.5.200
- 192.168.5.127
Correct answer: 192.168.5.190
The /26 subnet 192.168.5.128 covers .128–.191; only 192.168.5.190 falls in that range.
Question 7: How many subnets are created when you subnet 192.168.1.0/24 using a /28 mask?
- 16 (Correct answer)
- 14
- 8
- 32
Correct answer: 16
Borrowing 4 bits from a /24 (28 - 24 = 4) creates 2^4 = 16 subnets.
ICND1 Subnetting 2
How many usable host addresses are available in a /27 subnet?