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Exam N10-007 topic 1 question 231 discussion

Actual exam question from CompTIA's N10-007
Question #: 231
Topic #: 1
[All N10-007 Questions]

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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Suggested Answer: C 🗳️
by Jose0220 at Oct. 11, 2019, 12:31 a.m.

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Eye_Tea
Highly Voted 5 years, 2 months ago
THE ANSWER IS NOT "A". THE ANSWER IS "C", ALL DAY -- EVERY WAY! Engineering staff has 6 users. engineering subnet = 192.168.1.64/29; /29 means 3 host bits, block size = 8 thus subnet mask = 255.255.255.248 Network ID= 192.168.1.64 First host ip= 192.168.1.65 Last host ip= 192.168.1.70 Broadcast ID= 192.168.1.71 Next network address= 192.168.1.72 That provided answer is a joke
upvoted 26 times
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TriBiT
Highly Voted 5 years, 5 months ago
I think the answer should be C... if they are sub-netting even further 192.168.1.64/29 wold be the start of a new sub net with 8 address 6 usable after you take out the id and broadcast making it 192.168.1.70...the answer A just added 55+6 on the default sub-net for the remote office. the way i read the question it is asking for the engineering sub-net... am i wrong?
upvoted 14 times
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GingerWarrior69
Most Recent 3 years, 11 months ago
I still think the question is written really badly. Why would you mention the digital department when they don't ask you to calculate anything related to that department. Also why say using the least amount of IP's when they have already given you the CIDR notation anyway?
upvoted 1 times
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[Removed]
4 years, 2 months ago
The answer is C. 192.168.1.70. The engineering subnet is 192.168.1.64/29. Ip range for subnet is: 192.168.1.64-192.168.1.71 192.168.1.64 - subnet IP address 192.168.1.71 - broadcast IP address Usable IP addresses: 192.168.1.65-192.168.70 Last usable IP address is 192.168.1.70.
upvoted 2 times
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SarahSou
4 years, 3 months ago
This question make me wonder of other questions might be wrong...
upvoted 2 times
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bryan77
4 years, 4 months ago
"A"would be the answer is the question had been the last usable ip address for the digital marketing subnet.
upvoted 2 times
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xsp
4 years, 5 months ago
Answer is C,@ 192.168.1.64/29 leaves us with just 3 host bits since 29 is already dedicated for the network bits. Now 2^3 = 8 addresses - 2 (1 for the net id and 1 for the broadcast). .64 is the net id so the 1st usable add is .65 up to .70, .71 is reserved for the broadcast. Since the engineering team has 6 members then .65 ~ .70 will suffice. Then .70 would be the last usable add. for the engineering team. Cheers!
upvoted 1 times
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jay321
4 years, 8 months ago
Use a subnet calculator if not sure .. answer is C
upvoted 3 times
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Poppins
4 years, 11 months ago
So when I wrote it all out, I got 192.168.1.70 for the last valid host. Which is C.
upvoted 4 times
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Bingo
4 years, 11 months ago
Answer is C
upvoted 3 times
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AbiolaBad
5 years, 2 months ago
It's definitely C
upvoted 3 times
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elevator44
5 years, 2 months ago
CIDR calculator says : First IP 192.168.1.64, Last IP 192.168.1.71 so last usable host is minus one from last IP so C should be answer.
upvoted 5 times
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FransRSA
5 years, 3 months ago
Answer is C, as D's .71 will be the broadcast ip https://ip.rst.im/ip/192.168.1.64/29
upvoted 3 times
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PerkDizzzle
5 years, 3 months ago
I agree with answer being C.
upvoted 5 times
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Rz10
5 years, 3 months ago
I'm thinking, the sub is .64 , and you are adding another 8 ids, out of which 6 can be used. So, 64 is network id, and 72 is the broadcast id, then you have the last one .71 hence D w'd be correct.
upvoted 6 times
IhcimYmra
5 years, 2 months ago
.72 would be the network ID of the next SN (64 + 8 = 72), so .71 would be the BC ID and .70 would be the last usable host.
upvoted 7 times
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Pierrek
5 years, 3 months ago
The answer is C. A is not on the same subnet
upvoted 8 times
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