Question

you have a network address of 132.66.0.0 and a subnet mask of 255.255.224.0. which of the following are valid subnet addresses? (select two.) 132.98.0.0 132.130.0.0 132.66.255.0 132.66.96.0 132.66.192.0

Explanation:

Step1: Identify subnet mask bits

Subnet mask 255.255.224.0 → third octet is 224 ($11100000$ in binary). The subnet increment is $256 - 224 = 32$.

Step2: Check valid third octet values

Valid third octet values are multiples of 32 (0,32,64,96,128,160,192,224).

Step3: Verify each option

• 132.98.0.0: Third octet 98 → not multiple of 32.
• 132.130.0.0: Third octet 130 → not multiple of 32.
• 132.66.255.0: Third octet 255 → not multiple of 32.
• 132.66.96.0: Third octet 96 → multiple of 32.
• 132.66.192.0: Third octet 192 → multiple of 32.

Answer:

132.66.96.0, 132.66.192.0