ICND1 Subnetting 4 — Questions and Answers
Question 1: You need to divide 192.168.50.0/24 into subnets supporting 60 hosts each. Which mask should you use?
- /26 (Correct answer)
- /27
- /25
- /28
Correct answer: /26
A /26 provides 62 usable hosts (64 - 2), which is the smallest mask that supports 60 hosts.
Question 2: What is the network address of the subnet containing 172.16.35.200/20?
- 172.16.32.0 (Correct answer)
- 172.16.35.0
- 172.16.48.0
- 172.16.16.0
Correct answer: 172.16.32.0
/20 has a block size of 16 in the third octet; 35 falls in the 32–47 range, so the network is 172.16.32.0.
Question 3: Which of the following addresses is a valid host in the subnet 10.0.0.128/25?
- 10.0.0.200 (Correct answer)
- 10.0.0.127
- 10.0.0.255
- 10.0.1.1
Correct answer: 10.0.0.200
The subnet 10.0.0.128/25 covers 10.0.0.128–10.0.0.255; only 10.0.0.200 is a valid host (not the broadcast .255).
Question 4: How many bits must be borrowed from the host portion of a Class B address to create 100 subnets?
- 7 (Correct answer)
- 6
- 8
- 5
Correct answer: 7
2^7 = 128 ≥ 100 and 2^6 = 64 < 100, so 7 bits must be borrowed.
Question 5: What is the broadcast address of the subnet 10.20.30.64/27?
- 10.20.30.95 (Correct answer)
- 10.20.30.127
- 10.20.30.63
- 10.20.30.96
Correct answer: 10.20.30.95
A /27 has a block size of 32; the subnet starting at .64 ends at .64 + 31 = .95, which is the broadcast.
Question 6: Which decimal value does the binary octet 11111000 represent in a subnet mask?
- 248 (Correct answer)
- 252
- 240
- 254
Correct answer: 248
11111000 = 128+64+32+16+8 = 248.
Question 7: A /24 network is subnetted into /28 subnets. How many hosts can each subnet support?
- 14 (Correct answer)
- 16
- 30
- 12
Correct answer: 14
A /28 leaves 4 host bits; 2^4 - 2 = 14 usable host addresses per subnet.
ICND1 Subnetting 4
You need to divide 192.168.50.0/24 into subnets supporting 60 hosts each.
Which mask should you use?