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Exam N10-008 topic 1 question 85 discussion

Actual exam question from CompTIA's N10-008
Question #: 85
Topic #: 1
[All N10-008 Questions]

A workstation is configured with the following network details:

Software on the workstation needs to send a query to the local subnet broadcast address. To which of the following addresses should the software be configured to send the query?

  • A. 10.1.2.0
  • B. 10.1.2.1
  • C. 10.1.2.23
  • D. 10.1.2.255
  • E. 10.1.2.31
Show Suggested Answer Hide Answer
Suggested Answer: E 🗳️

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Lu5ck
Highly Voted 2 years, 3 months ago
Selected Answer: E
CIDR 27 is 1111 1111.1111 1111.1111 1111.1110 0000 2^5 = 32 hosts per subnet 1st address is network address, last address is broadcast address The user address is 10.1.2.23 which means he reside in the network of 10.1.2.0 to 10.1.2.31 10.1.2.31 is the last address in that network thus the broadcast address
upvoted 55 times
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ExamTopicsDiscussor
Highly Voted 1 year, 9 months ago
Easier to use sunny's method, 1 2 4 8 16 32 64 128 256 256 128 64 32 16 8 4 2 1 24 25 26 27 28 29 30 31
upvoted 13 times
MacheenZero
1 year, 5 months ago
For those that don't know what this is https://www.youtube.com/watch?v=ecCuyq-Wprc&ab_channel=SunnyClassroom
upvoted 18 times
user82
1 year, 2 months ago
Thanks for the link. Can someone explain how the first subnet is 10.1.2.0 - 10.1.2.31 if the default gateway is 10.1.2.1? Is that an error? I feel like the default gateway should be 10.1.2.0
upvoted 2 times
famco
1 year, 2 months ago
You should do the subnetting chapters properly. 10.1.2.0 is the network Id. That is not an assignable IP address. Default gateway is an IP address in the same network and it cannot be the network Id.
upvoted 3 times
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LiamAzure
Most Recent 10 months ago
/27 is 255.255.255.224 256-224=32 The subnets go up by 32 10.1.2.0-10.1.2.31 is the first subnet, the first and last addresses are unusable and reserved for broadcast address
upvoted 12 times
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famco
1 year, 2 months ago
Don't make it complex. The first 3 bits of the last octet is used for the subnet leaving 5 bits for the host portion. So it could be from 0-31, 32-63, 64- 95, etc 2^3 = 8 subnets like that. Luckily only have to look at the first. and last in that range (31) is the broadcast (first is the network id).
upvoted 3 times
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S8766
2 years, 2 months ago
Selected Answer: E
10.1.2.31 is the last address in that network
upvoted 3 times
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IAmTheSenate6699
2 years, 3 months ago
Selected Answer: E
Voting E .31 would be broadcast address of the subnet
upvoted 2 times
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waqdhiyo
2 years, 4 months ago
Selected Answer: B
It should send to the default gateway IP.
upvoted 1 times
RobJob
2 years, 3 months ago
The question is asking the IP for a broadcast address not a default gateway. A broadcast address and default gateway are not the same thing.
upvoted 7 times
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ebtesam
2 years, 5 months ago
can someone explain why E , plz
upvoted 3 times
Pongsathorn
2 years, 5 months ago
/27 have 32 addresses per subnet 1st subnet start with 10.1.2.0 - 10.1.2.31 2nd subnet start with 10.1.2.32 - 10.1.2.63 3rd subnet start with 10.1.2.64 - 10.1.2.95 And so on. IP address 10.1.2.23 is in the 1st subnet. So, broadcast address is 10.1.2.31
upvoted 26 times
user82
1 year, 2 months ago
How is the first subnet 10.1.2.0 - 10.1.2.31 if the default gateway is 10.1.2.1 ?
upvoted 2 times
TheFivePips
10 months, 2 weeks ago
If I understand this correctly, the x.x.x.0 is the network id for this first subnet (the one that the workstation is on), and is not used for hosts. The x.x.x.31 is the broadcast id; also not used for hosts. The default gateway is often, but does NOT have to be, the first AVAILABLE host of the subnet (which in this case is x.x.x.1)
upvoted 1 times
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l3fty
2 years, 4 months ago
Thanks for this explanation, I was confused as well.
upvoted 5 times
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