An interface is configured with 10.24.7.158/27. What is the broadcast address of that subnet?
Correct. It is the last address in the /27 block.
Why this answer
A /27 uses blocks of 32 addresses. The block containing .158 is 10.24.7.128 through 10.24.7.159, so .159 is the broadcast address.
Exam trap
A frequent exam trap is mistaking the broadcast address for the network address of the next subnet or the last address of a different subnet block. For example, 10.24.7.160 is the network address of the next /27 subnet, not the broadcast address of the current subnet. Candidates often confuse the last usable host address with the broadcast address or select an address from an adjacent subnet block.
This mistake leads to incorrect subnetting answers and can cause interface misconfigurations in real networks, resulting in communication failures.
Why the other options are wrong
Incorrect. 10.24.7.191 is the last address of a larger /26 subnet block, not the /27 block containing 10.24.7.158, so it cannot be the broadcast address here.
Incorrect. 10.24.7.127 is the broadcast address of the previous /27 subnet block (10.24.7.96/27), not the one containing 10.24.7.158.
Incorrect. 10.24.7.160 is the network address of the next /27 subnet block after 10.24.7.128/27, not the broadcast address of the current subnet.