The address of a class B host is to be split into subnets with a 6-bit subnet…

Concept:

Classical subnetting splits a network's host-bit space into a subnet field (s bits) and a host field (h bits). Under the classical convention that many exam keys still follow, the all-zeros and all-ones bit patterns are excluded from BOTH fields — a subnet-field value of all 0s or all 1s is not permitted, and within each subnet the all-zeros host address is that subnet's network address while the all-ones host address is its broadcast address. So the usable count in a field of n bits is 2n − 2, not 2n itself.

Application:

1. A Class B address has a default /16 mask, so 16 bits are available for subnetting and host addressing combined.

2. Borrowing 6 bits for the subnet field leaves 16 − 6 = 10 bits for host addressing inside each subnet.

3. Usable subnets = 26 − 2 = 64 − 2 = 62 (excluding the all-zeros and all-ones subnet-field patterns).

4. Usable hosts per subnet = 210 − 2 = 1024 − 2 = 1022 (excluding the network and broadcast addresses of each subnet).

Cross-check:

Borrowing 6 subnet bits turns the /16 Class B mask into a /22 mask (16 + 6 = 22). A /22 mask leaves 32 − 22 = 10 host bits, independently confirming 1022 usable hosts per subnet.

62 subnets and 1022 hosts per subnet is the value matching this exam's official key. Note: under the more modern convention (RFC 1878), the all-zeros/all-ones subnets are also treated as usable, which would give 64 usable subnets while the host count stays at 1022 — but that exact 64-and-1022 pairing is not among the options offered here, and this exam's official key follows the classical convention applied above.