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Exam N10-008 topic 1 question 327 discussion

Actual exam question from CompTIA's N10-008
Question #: 327
Topic #: 1
[All N10-008 Questions]

A technician is configuring a static IP address on a new device in a newly created subnet. The work order specifies the following requirements:

• The IP address should use the highest address available in the subnet.
• The default gateway needs to be set to 172.28.85.94.
• The subnet mask needs to be 255.255.255.224.

Which of the following addresses should the engineer apply to the device?

  • A. 172.28.85.93
  • B. 172.28.85.95
  • C. 172.28.85.254
  • D. 172.28.85.255
Show Suggested Answer Hide Answer
Suggested Answer: A 🗳️
by AustinKelleyNet at Dec. 10, 2022, 10:29 p.m.

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PaytoPlay
Highly Voted 1 year, 2 months ago
Selected Answer: A
Here is how I get my answer: 255.255.255.224 /27 256 - 224= Block Size 32 Block Size Subnet Range: 172.28.85.0 172.28.85.32 172.28.85.64 172.28.85.96 172.28.85.128 Broadcast Range: -1 the Block size 172.28.85.31 172.28.85.63 172.28.85.95 172.28.85.127 Usable IP Range: 172.28.85.1 - 172.28.85.30 172.28.85.33 - 172.28.85.62 172.28.85.65 - 172.28.85.94 172.28.85.97 - 172.28.85.126
upvoted 31 times
Mehsotopes
7 months ago
Default gateway might be setting up where the IP address can reach up to, otherwise usable IP range having -2 from the last address doesn't make sense. The first part of an IP address is used as a network address, the last part as a host address. If you take the example 192.168.123.132 and divide it into these two parts, you get 192.168.123. Network .132 Host or 192.168.123.0 - network address. 0.0.0.132 - host address. https://learn.microsoft.com/en-us/troubleshoot/windows-client/networking/tcpip-addressing-and-subnetting#:~:text=The%20first%20part%20of%20an,0.0.0.132%20%2D%20host%20address.
upvoted 2 times
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famco
1 year, 2 months ago
That's an easier way for this question. Thanks
upvoted 4 times
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Cohort07
Highly Voted 1 year, 2 months ago
Selected Answer: A
The answer is A. The subnet ranges for /27 are: 0-31 32-63 64-95 96-127 etc... Normally the default gateway is the first usable IP in that subnet range however in this question it is setting the default gateway to 172.28.85.94. This means that 172.28.85.94 and 172.28.85.95 are the default gateway and broadcast addresses. That means the only correct IP from the answer choices is A. 172.28.85.93.
upvoted 10 times
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RobV
Most Recent 1 year, 2 months ago
Selected Answer: B
The subnet mask 255.255.255.224 indicates that this is a /27 subnet, which allows for 32 IP addresses (30 usable addresses after reserving the network and broadcast addresses). The highest address available in this subnet is 172.28.85.94 (which is the default gateway), so we need to use the next available IP address. Therefore, the correct IP address to assign to the device is 172.28.85.95, which is option B. Option A (172.28.85.93) is the address immediately preceding the default gateway, which is not the highest address available in the subnet. Option C (172.28.85.254) and option D (172.28.85.255) are outside the range of usable addresses in this subnet, as they are the network and broadcast addresses, respectively.
upvoted 2 times
kerppa
1 year ago
172.28.85.95 is reserved for the broadcast address, so the answer is A
upvoted 2 times
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[Removed]
1 year, 3 months ago
Im sorry but how is it not B? The default gateway is the first IP address used in the acceptable range, broadcast is the last. So the next available IP address after .94 is .95
upvoted 1 times
famco
1 year, 2 months ago
95 is the broadcast address which is the last. The next range starts with 96
upvoted 2 times
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max319
1 year, 4 months ago
Correct me if I'm wrong, but I believe it is B. Considering the IP's assigned give us 30 nodes with the reserved network address being .93 and broadcast being .125. The gateway is typically put on the first IP being .94 which it is. Then the first available host address would be .95. The only reason I am not fully confident is because I don't know whether or not the IP range starts on .93 or .64.
upvoted 3 times
ghosting
1 year, 4 months ago
The broadcast is the .95 The gateway is the .94 Last available IP is .93
upvoted 5 times
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StrawberryTechie
9 months, 3 weeks ago
the last IP address in every subnet is always the broadcast address. The first is always the network address. .94 is used for the default gateway. So .93 is the last available. A is correct.
upvoted 1 times
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famco
1 year, 2 months ago
There is no shortcut than to understand the calculation.
upvoted 1 times
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JakeCharles
1 year, 4 months ago
Selected Answer: A
just checked online , A should be correct
upvoted 1 times
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TomSawyer
1 year, 6 months ago
https://www.tunnelsup.com/subnet-calculator/ IP Address: 172.28.85.95/27 Netmask: 255.255.255.224 Network Address: 172.28.85.64 Usable Host Range: 172.28.85.65 - 172.28.85.94 Broadcast Address: 172.28.85.95
upvoted 2 times
[Removed]
1 year, 3 months ago
How did you get the IP address though? To be specific how did you get .95?
upvoted 2 times
ep0ch
1 year, 2 months ago
The broadcast address is always the last in the subnet range. With the gateway being set to .94, the highest available address left is .93
upvoted 2 times
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AustinKelleyNet
1 year, 6 months ago
Selected Answer: A
This is correct.
upvoted 1 times
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Community vote distribution
A (35%)
C (25%)
B (20%)
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