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Exam 200-150 topic 1 question 26 discussion

Actual exam question from Cisco's 200-150
Question #: 26
Topic #: 1
[All 200-150 Questions]

Which two options represent a subnet mask that allows for a maximum of 14 available host addresses on a subnet? (Choose two.)

  • A. 255.255.255.240
  • B. 255.255.255.192
  • C. 255.255.240.0
  • D. 255.255.192.0
  • E. /14
  • F. /21
  • G. /26
  • H. /28
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Suggested Answer: AH 🗳️
Example:

Address: 192.168.1.1 -

Bitmask: 28 -

Netmask: 255.255.255.240 -

Network: 192.168.1.0 -

HostMin: 192.168.1.1 -

HostMax: 192.168.1.14 -

Broadcast: 192.168.1.15 -
Hosts: 16
by snapper at Aug. 22, 2019, 10:06 p.m.

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snapper
5 years, 10 months ago
Remember the formula: 2^n - 2 where "n" is the number of hosts bits (and you subtract one for the network address and one for the broadcast address). So, 2x2x2x2 (2^4) = 16. Subtract 2 (network and broadcast address) = 14 We're only borrowing host bits from the last octet of the mask, so the mask is: 11111111.11111111.11111111.1111xxxx The "x"s are the host bits we borrowed. The "ones" represent the subnet. 255.255.255.240 (the ones position equate to 128+64+32+16 = 240).
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