IPv4 Addressing & Subnetting Practice
Exam 2026/2027 | 100 Questions with
Answers & Explanations | CIDR, VLSM,
Private Ranges, Broadcast Addresses
Description:
Master IPv4 subnetting for 2026/2027 with 100 certification-style questions. Covers CIDR,
VLSM, host calculations, private address ranges, and broadcast identification. Full answers
and detailed explanations included.
Download now to pass your networking exam on the first attempt.
, IPv4 Subnetting Exam 2026/2027 – 100 Q&A
Academic Year: 2026/2027
Level: Advanced Networking / University Foundation
Total Questions: 100
Format: Multiple Choice, Fill-in-the-Blank, Multiple Answer
Section 1: Subnet Masks and CIDR Notation
1. What is the dot-decimal representation of a /13 subnet mask?
A. 255.240.0.0
B. 255.248.0.0
C. 255.252.0.0
D. 255.254.0.0
Answer: B
Explanation: A /13 subnet mask indicates that the first 13 bits are set to 1. The first octet (8 bits)
is 255, and the second octet has the first 5 bits set to 1 (since 13 - 8 = 5), which equals 248 in
decimal. The remaining octets are 0, resulting in 255.248.0.0.
2. Which of the following answers lists the CIDR notation for the 255.255.224.0 subnet mask?
A. /18
B. /19
C. /20
D. /21
Answer: B
Explanation: The mask 255.255.224.0 has 8+8+3 = 19 bits set to 1 (since 224 = 128+64+32).
The CIDR notation is therefore /19.
,3. Which of the following answers lists the CIDR notation of the 255.255.255.224 subnet mask?
A. /25
B. /26
C. /27
D. /28
Answer: C
Explanation: The mask 255.255.255.224 has 24 bits from the first three octets plus 3 bits from
the last octet (224 = 128+64+32), totaling 27 bits. The correct CIDR notation is /27.
4. Which of the following answers lists the CIDR notation of the 255.192.0.0 subnet mask?
A. /8
B. /9
C. /10
D. /11
Answer: C
Explanation: The first octet (255) contributes 8 bits. The second octet (192) contributes 2 bits
(since 192 = 128+64). The total is 8 + 2 = 10 bits, so the CIDR notation is /10.
5. What is the binary representation of the 255.255.128.0 subnet mask?
A. 11111111.11111111.10000000.00000000
B. 11111111.11111111.00000000.00000000
C. 11111111.11111111.11000000.00000000
D. 11111111.11111110.00000000.00000000
Answer: A
Explanation: 255 in binary is 11111111. 128 in binary is 10000000. The mask 255.255.128.0
converts to 11111111.11111111.10000000.00000000.
, 6. Which of the following answers lists the binary notation of the decimal number 252?
A. 11111000
B. 11111100
C. 11111110
D. 11111111
Answer: B
Explanation: 252 in binary is 11111100. This is derived from 128+64+32+16+8+4 = 252, with
the last two bits as 0.
7. Which of the following is an example of a valid subnet mask?
A. 255.255.255.256
B. 255.0.255.0
C. 0.255.255.255
D. None of the above is a valid subnet mask
Answer: D
Explanation: A valid subnet mask consists of consecutive 1s starting from the leftmost bit,
followed by consecutive 0s. Option A is invalid because 256 exceeds 255. Option B
(255.0.255.0) has non-consecutive 1s. Option C (0.255.255.255) does not start with 1s on the
left. None are valid.
Section 2: Network Addressing and Host Calculations
8. How many usable IP addresses can be assigned to hosts on a /26 subnet?
A. 30
B. 62
C. 64
D. 126
Exam 2026/2027 | 100 Questions with
Answers & Explanations | CIDR, VLSM,
Private Ranges, Broadcast Addresses
Description:
Master IPv4 subnetting for 2026/2027 with 100 certification-style questions. Covers CIDR,
VLSM, host calculations, private address ranges, and broadcast identification. Full answers
and detailed explanations included.
Download now to pass your networking exam on the first attempt.
, IPv4 Subnetting Exam 2026/2027 – 100 Q&A
Academic Year: 2026/2027
Level: Advanced Networking / University Foundation
Total Questions: 100
Format: Multiple Choice, Fill-in-the-Blank, Multiple Answer
Section 1: Subnet Masks and CIDR Notation
1. What is the dot-decimal representation of a /13 subnet mask?
A. 255.240.0.0
B. 255.248.0.0
C. 255.252.0.0
D. 255.254.0.0
Answer: B
Explanation: A /13 subnet mask indicates that the first 13 bits are set to 1. The first octet (8 bits)
is 255, and the second octet has the first 5 bits set to 1 (since 13 - 8 = 5), which equals 248 in
decimal. The remaining octets are 0, resulting in 255.248.0.0.
2. Which of the following answers lists the CIDR notation for the 255.255.224.0 subnet mask?
A. /18
B. /19
C. /20
D. /21
Answer: B
Explanation: The mask 255.255.224.0 has 8+8+3 = 19 bits set to 1 (since 224 = 128+64+32).
The CIDR notation is therefore /19.
,3. Which of the following answers lists the CIDR notation of the 255.255.255.224 subnet mask?
A. /25
B. /26
C. /27
D. /28
Answer: C
Explanation: The mask 255.255.255.224 has 24 bits from the first three octets plus 3 bits from
the last octet (224 = 128+64+32), totaling 27 bits. The correct CIDR notation is /27.
4. Which of the following answers lists the CIDR notation of the 255.192.0.0 subnet mask?
A. /8
B. /9
C. /10
D. /11
Answer: C
Explanation: The first octet (255) contributes 8 bits. The second octet (192) contributes 2 bits
(since 192 = 128+64). The total is 8 + 2 = 10 bits, so the CIDR notation is /10.
5. What is the binary representation of the 255.255.128.0 subnet mask?
A. 11111111.11111111.10000000.00000000
B. 11111111.11111111.00000000.00000000
C. 11111111.11111111.11000000.00000000
D. 11111111.11111110.00000000.00000000
Answer: A
Explanation: 255 in binary is 11111111. 128 in binary is 10000000. The mask 255.255.128.0
converts to 11111111.11111111.10000000.00000000.
, 6. Which of the following answers lists the binary notation of the decimal number 252?
A. 11111000
B. 11111100
C. 11111110
D. 11111111
Answer: B
Explanation: 252 in binary is 11111100. This is derived from 128+64+32+16+8+4 = 252, with
the last two bits as 0.
7. Which of the following is an example of a valid subnet mask?
A. 255.255.255.256
B. 255.0.255.0
C. 0.255.255.255
D. None of the above is a valid subnet mask
Answer: D
Explanation: A valid subnet mask consists of consecutive 1s starting from the leftmost bit,
followed by consecutive 0s. Option A is invalid because 256 exceeds 255. Option B
(255.0.255.0) has non-consecutive 1s. Option C (0.255.255.255) does not start with 1s on the
left. None are valid.
Section 2: Network Addressing and Host Calculations
8. How many usable IP addresses can be assigned to hosts on a /26 subnet?
A. 30
B. 62
C. 64
D. 126
