Question 70
A host uses 172.17.58.242/26 as its IP address and needs to ARP for its gateway. Which proposed gateway is in the same /26 and is an ordinary host?
Explanation
Correct Answer
With /26, addresses .192 through .255 form one subnet, with .192 network and .255 broadcast. Both .242 and .193 are host values in that shared block, so the endpoint can resolve the gateway directly over the local VLAN. A gateway in the preceding /26 would need routing and is not directly on-link.
Incorrect Answers
The .191 address ends the previous /26 block .128–.191 and is its broadcast. It is not an eligible gateway on the host’s .192/26 subnet.
The .192 address begins the required subnet but is the network identifier, not a conventional usable interface address. The gateway must use an eligible host ID.
That address is outside the 172.17.58.192/26 block and cannot be directly ARPed on the stated subnet. It would require a separate network path.