Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
A . 192.168.1.8/29
B . 192.168.1.32/27
C . 192.168.1.40/29
D . 192.168.1.16/28
E . 192.168.1.48/29
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
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