|A) 255.255.0.0||B) 255.0.0.0|
|C) 255.255.240.0||D) 255.255.254.0|
Explanation: The requirement of having to allocate 500 hosts on each subnet can be achieved by the following: 2^9 = 512. So deducting the broadcast and network address we have 510 possible hosts on each subnet. We used 9 bits to allocate the hosts, this leaves us 7 bits for possible subnets. Those 7 bits in binary is 1111 1110 or 254 in decimal.