The subnet mask for a particular network is 255.255.31.0. Which of the…
2009
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?
- A.
172.57.88.62 and 172.56.87.23
- B.
10.35.28.2 and 10.35.29.4
- C.
191.203.31.87 and 191.234.31.88
- D.
128.8.129.43 and 128.8.161.55
Attempted by 108 students.
Show answer & explanation
Correct answer: D
To determine if two IP addresses belong to the same network, they must have the exact same Network ID. The Network ID is calculated by performing a bitwise AND operation between the IP address and the Subnet Mask.
1. Analyze the Subnet Mask
The mask is 255.255.31.0.
1st and 2nd Octets (255.255): These are fixed. For two IPs to be in the same network, their first two octets must match exactly.
3rd Octet (31): This is the "interesting" octet. In binary, 31 is
00011111.4th Octet (0): This is part of the host portion and does not affect the Network ID.
2. Evaluate the Options
Option A: 172.57.88.62 and 172.56.87.23
The second octets are 57 and 56.
Since the mask is 255.255.x.x, the first two octets must be identical. These are different.
Result: Different Networks.
Option B: 10.35.28.2 and 10.35.29.4
The first two octets match (10.35). Let's check the 3rd octet (28 and 29) against the mask (31).
Binary 28:
00011100Binary 29:
00011101Binary Mask 31:
00011111AND Operation:
28 AND 31 = 28
29 AND 31 = 29
The Network IDs are different (10.35.28.0 vs 10.35.29.0).
Result: Different Networks.
Option C: 191.203.31.87 and 191.234.31.88
The second octets are 203 and 234.
Since the mask for the second octet is 255, these must match. They do not.
Result: Different Networks.
Option D: 128.8.129.43 and 128.8.161.55
The first two octets match (128.8). Let's check the 3rd octet (129 and 161) against the mask (31).
Binary 129:
10000001Binary 161:
10100001Binary Mask 31:
00011111AND Operation:
129 AND 31:
10000001AND00011111=00000001(1)161 AND 31:
10100001AND00011111=00000001(1)
Both addresses result in the same Network ID: 128.8.1.0.
Final Answer
The correct pair is 128.8.129.43 and 128.8.161.55.
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