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Question ID 6383

Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?

Option A

192.168.1.8/29

Option B

192.168.1.32/27

Option C

192.168.1.40/29

Option D

192.168.1.16/28

Option E

192.168.1.48/29

Correct Answer c
Explanation 248 mask uses 5 bits (1111 1000) 42 IP in binary is (0010 1010) The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP... 1111 1000 AND 0010 1010 equals 0010 1000 - which is .40 /24 is standard class C mask. Adding the 5 bits from the .248 mask gives /29


Question ID 6384

Which two statements describe the IP address 10.16.3.65/23? (Choose two.)

Option A

The subnet address is 10.16.3.0 255.255.254.0.

Option B

The lowest host address in the subnet is 10.16.2.1 255.255.254.0.

Option C

The last valid host address in the subnet is 10.16.2.254 255.255.254.0

Option D

The broadcast address of the subnet is 10.16.3.255 255.255.254.0.

Option E

The network is not subnetted.

Correct Answer bd
Explanation

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