You have subnetted a Class B address space of 1379900 using

You have subnetted a Class B address space of 137.99.0.0 using a subnet mask of 255.255.252.0. What is the broadcast address of the third subnet in your design?

A. 137.99.12.255

B. 137.99.15.255

C. 137.99.8.255

D. 137.99.11.255

Solution

D is correct. The 255.255.252.0 subnet mask for 137.99.0.0 borrows 6 bits from the host field. The third combination of these bits equates to the “8” subnet (000000|xx is the first, 000001|xx is the second, and 000010|xx is the third, where the x’s represent the 2 host bits in the third octet). To find the broadcast address, all host bits must be set to 1s: 000010|11.11111111. This equates to 137.99.11.255.

A, B, and C are incorrect. A is not correct because since all host bits are not set to 1s, 137.99.12.255 is simply a host address on the fourth subnet (000011|00.11111111). B is not correct because 137.99.15.255 is the broadcast address for the fourth subnet (000011|11.11111111). C is not correct because since all host bits are not set to 1s, 137.99.8.255 is a host address on the third octet (000010|00.11111111).