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Exam 102-500 All Questions

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Exam 102-500 topic 1 question 12 discussion

Actual exam question from LPI's 102-500
Question #: 12
Topic #: 1
[All 102-500 Questions]

How many IP addresses can be used for unique hosts inside the IPv4 subnet 192.168.2.128/26?

  • A. 6
  • B. 14
  • C. 30
  • D. 62
  • E. 126
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Suggested Answer: D 🗳️

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Jodelo
Highly Voted 2 years, 2 months ago
Each subnet mask has 32 bits which are sperate in network and host bits. In this case we have 28 network bits which means 32 - 28 = 6 Host bits remain. 2^6 = 64.. BUT every subnet has 1 network adress (the first IP address in that subnet) and 1 broadcast address (the last one in that subnet) so 64 -2 = 62 hosts in each subnet
upvoted 10 times
Jodelo
2 years, 2 months ago
EDIT: Sorry it's 32 - 26 of course
upvoted 2 times
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CuriousLinuxCat
Highly Voted 2 years, 1 month ago
Correct. IP for this subnet is 192.168.2.128 ~ .191. However, you need to remove the subnet address (.128) and broadcast address (.191) ...and you are left with 62 unique host addresses.
upvoted 8 times
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dfdfsdfdsfdf
Most Recent 8 months, 2 weeks ago
correct
upvoted 1 times
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ellen_AA
2 years, 6 months ago
2^(32-26) - 2 = 62 is the formula to determine the number of hosts available on a given subnet.
upvoted 5 times
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phucpeta
2 years, 7 months ago
2^6-2 because 2 ip for broadcast and subnet
upvoted 1 times
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Audio00_02
2 years, 8 months ago
ambecilli
upvoted 3 times
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Rini_Giannenzo
2 years, 9 months ago
Ma fateve na pera
upvoted 2 times
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Malvada
2 years, 9 months ago
D is correct, 64 available, but network and broadcast address are not available.
upvoted 3 times
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rjlg2centos8
2 years, 10 months ago
64 no, because the broadcast address and the net address doesn't count
upvoted 1 times
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Stefany
2 years, 10 months ago
correct B
upvoted 1 times
Stefany
2 years, 10 months ago
Perdon 64
upvoted 2 times
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C (25%)
B (20%)
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