A technician is allocating the IP address space needed for a new remote office. This office will contain the engineering staff with six employees and the digital marketing staff with 55 employees. The technician has decided to allocate the 192.168.1.0/24 block to the remote office. The engineering staff has been allocated the 192.168.1.64/29 subnet. Using the LEAST amount of space possible, which of the following would be the last usable IP address in the engineering subnet?
A. 192.168.1.62
B. 192.168.1.63
C. 192.168.1.70
D. 192.168.1.71
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192.168.1.64/29 = 8 BS
192.168.1.72/29
# of Usable host = 8 – 2 = 6
First Usable IP = x.x.x.65
Last Usable IP = x.x.x.70
Broadcast IP = x.x.x.71
answer here is C
Its absolutly 100% without doubt C.
/29 has 8 possible hosts. 0-7, 8-15, 16-23……….64-71.
71 is the broadcast address so 70 is the last usable.
Youre welcome 🙂
The correct answer is C because the question says what is the last usable IP in the engineering subnet
29/ = 8 IPs* lol
sayyyy whaaaaa
Look mane, I just counted on my hands /29 having 64 IPs. So that is 2 IPs to subtract. 1 for the Network and 1 for the Broadcast.
I was thinking .62 (It’s what’s in between that’s usable b)!
0; 64; 128…
63; 127;
the correct answer should be 192.168.1.70 = C not A!
engineers got
192.168.1.64/29
Network: 192.168.1.64/29 11000000.10101000.00000001.01000 000 (Class C)
Broadcast: 192.168.1.71 11000000.10101000.00000001.01000 111
HostMin: 192.168.1.65 11000000.10101000.00000001.01000 001
HostMax: 192.168.1.70 11000000.10101000.00000001.01000 110
Hosts/Net: 6 engineers
Agreed
there is 2 correct answers here, the host is multiple of 8 so the broadcast also can be .63 and the last usable ip .62
Sorry Sam, but the subnet is 64. and not the 56.That is, from 64 to 72.. so last ip add. is 70. 62 will be the last ip if the subnet was 56.