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?

  1. A.

    172.57.88.62 and 172.56.87.23

  2. B.

    10.35.28.2 and 10.35.29.4

  3. C.

    191.203.31.87 and 191.234.31.88

  4. D.

    128.8.129.43 and 128.8.161.55

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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: 00011100

  • Binary 29: 00011101

  • Binary Mask 31: 00011111

  • AND 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: 10000001

  • Binary 161: 10100001

  • Binary Mask 31: 00011111

  • AND Operation:

    • 129 AND 31: 10000001 AND 00011111 = 00000001 (1)

    • 161 AND 31: 10100001 AND 00011111 = 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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