What is the result of ANDing the IP address 23.176.224.18 with the subnet mask 255.224.0.0?

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Multiple Choice

What is the result of ANDing the IP address 23.176.224.18 with the subnet mask 255.224.0.0?

Explanation:
To determine the result of ANDing the IP address 23.176.224.18 with the subnet mask 255.224.0.0, we need to perform a bitwise AND operation between the two values. First, convert both the IP address and the subnet mask into binary format: The IP address 23.176.224.18 in binary is: - 23: 00010111 - 176: 10110000 - 224: 11100000 - 18: 00010010 Putting this together: 00010111.10110000.11100000.00010010 The subnet mask 255.224.0.0 in binary is: - 255: 11111111 - 224: 11100000 - 0: 00000000 - 0: 00000000 Putting this together: 11111111.11100000.00000000.00000000 Now, perform the AND operation bit by bit: 1. The first octet: 00010111 (23) AND 11111111 (255) = 00010111 (23) 2. The second octet

To determine the result of ANDing the IP address 23.176.224.18 with the subnet mask 255.224.0.0, we need to perform a bitwise AND operation between the two values.

First, convert both the IP address and the subnet mask into binary format:

The IP address 23.176.224.18 in binary is:

  • 23: 00010111

  • 176: 10110000

  • 224: 11100000

  • 18: 00010010

Putting this together:

00010111.10110000.11100000.00010010

The subnet mask 255.224.0.0 in binary is:

  • 255: 11111111

  • 224: 11100000

  • 0: 00000000

  • 0: 00000000

Putting this together:

11111111.11100000.00000000.00000000

Now, perform the AND operation bit by bit:

  1. The first octet:

00010111 (23)

AND 11111111 (255)

= 00010111 (23)

  1. The second octet
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